NCERT Solutions Class 12 Physics Chapter 7

Q.1 A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?

Ans.

$\begin{array}{l}\text{Here, resistance of the resistor, R=100 Ω}\\ \text{Supply voltage, V=220 V}\\ \text{Frequency,}\mathrm{\nu }\text{= 50 Hz}\\ \left(\text{a}\right)\text{The rms value of current in the circuit is given by,}\\ \text{I =}\frac{\text{V}}{\text{R}}\text{=}\frac{\text{220 V}}{\text{100 Ω}}\\ \text{= 2.20 A}\\ \left(\text{b}\right)\text{The net power consumed over a full cycle is given by,}\\ \text{P = VI}\\ \text{= 220 V × 2.2 A}\\ \text{= 484 W}\end{array}$

Q.2 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{Here,\hspace{0.33em}peak voltage of the ac supply},{\text{V}}_0=\text{3}00\text{V}\\ \text{Rms voltage is given by,}\\ \text{V=}\frac{{\text{V}}_0}{\sqrt{2}}\\ =\frac{300\mathrm{V}}{\sqrt{2}}=212.1\mathrm{V}\\ \left(\text{b}\right)\text{The\hspace{0.33em}rms value of current is given by,}\\ \text{I}=\text{1}0\text{A}\\ \therefore {\text{Peak current,\hspace{0.33em}I}}_{\text{0}}=\sqrt{2}\mathrm{I}=\text{1}0\sqrt{2}\mathrm{A}=14.1\mathrm{A}\end{array}$

Q.3 A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

Ans.

$\begin{array}{l}{\text{Here,\hspace{0.33em}inductance of inductor, L = 44 mH = 44 × 10}}^{\text{-3}}\text{H}\\ \text{Supply voltage, V = 220 V}\\ \text{Frequency,}\mathrm{\nu }\text{= 50 Hz}\\ \text{Angular frequency, ω = 2}\mathrm{\pi \nu }\to \text{(i)}\\ {\text{Inductive reactance, X}}_{\text{L}}\text{= ω L}\to \text{(ii)}\\ \text{From\hspace{0.33em}equation\hspace{0.33em}(i)\hspace{0.33em}and\hspace{0.33em}(ii),\hspace{0.33em}we\hspace{0.33em}have}\\ {\text{X}}_{\text{L}}\text{= 2}\mathrm{\pi \nu }\text{L}\\ {\text{X}}_{\text{L}}\text{= 2}\mathrm{\pi }{\text{× 50 Hz × 44 ×10}}^{\text{-3}}\text{\hspace{0.33em}Ω}\\ \text{Rms value of current,\hspace{0.33em}I =}\frac{\text{V}}{{\text{X}}_{\text{L}}}\\ \text{I =}\frac{\text{220 V}}{{\text{2π×50 Hz × 44×10}}^{\text{-3}}\text{Ω}}\text{=15.92\hspace{0.33em}A}\\ \therefore \text{The rms value of current in the circuit is 15.92 A.}\end{array}$

Q.4 A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

Ans.

$\begin{array}{l}{\text{Here,\hspace{0.33em}capacitance of capacitor, C = 60 μF = 60×10}}^{\text{-6}}\text{F}\\ \text{Supply voltage, V = 110 V}\\ \text{Frequency,}\mathrm{\nu }\text{= 60 Hz}\\ \text{Angular frequency, ω = 2}\mathrm{\pi \nu }\\ \text{Capacitive reactance\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}by,}\\ {\text{X}}_{\text{c}}\text{=}\frac{\text{1}}{\text{ωC}}\text{=}\frac{\text{1}}{\text{2}\mathrm{\pi \nu }\text{C}}\\ \text{=}\frac{\text{1}}{{\text{2× 3.14 × 60 Hz × 60 ×10}}^{\text{-6}}\mathrm{F}}{\text{\hspace{0.17em}\hspace{0.17em}Ω}}^{\text{-1}}\\ \text{Rms value of current is\hspace{0.33em}given\hspace{0.33em}by,\hspace{0.33em}}\\ \text{I =}\frac{\text{ν}}{{\text{X}}_{\text{c}}}\\ {\text{= 110 × 2 × 3.14 × 60 Hz × 60 × 10}}^{\text{-6}}\mathrm{F}\\ \text{= 2.49\hspace{0.33em}A}\\ \therefore \text{The rms value of current is 2.49 A.}\end{array}$

Q.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

Ans.

$\begin{array}{l}\text{In Exercise\hspace{0.33em}7.3,}\\ \text{Rms value of the\hspace{0.33em}current, I = 15.92 A}\\ \text{Rms value of the\hspace{0.33em}voltage, V = 220 V}\\ \text{The net power absorbed can be calculated by the relation,}\\ \text{P = VI cos}\mathrm{\varphi }\\ \text{Here,\hspace{0.17em}\hspace{0.17em}}\mathrm{\varphi }\text{= Phase difference between V and I}\\ \text{In a pure inductive circuit, the phase difference}\\ \text{between alternating voltage and current\hspace{0.33em}is\hspace{0.33em}90° i.e.,}\\ \mathrm{\varphi }\text{= 90°}\\ \therefore \text{P = VI cos 90° = 0}\\ \therefore \text{The net power is zero.}\\ \text{In Exercise\hspace{0.33em}7.4,}\\ \text{Rms value of the\hspace{0.33em}current,}\\ \text{I = 2.49 A}\\ \text{Rms value of the\hspace{0.33em}voltage,}\\ \text{V = 110 V}\\ \text{The net power absorbed can be obtained as:}\\ \text{P = VI cos}\mathrm{\varphi }\\ \text{In a pure capacitive circuit, the phase difference}\\ \text{between alternating voltage and\hspace{0.33em}current is 90° i.e.,}\\ \mathrm{\varphi }\text{= 90°.}\\ \therefore \text{P = VI cos 90° = 0}\\ \therefore \text{The net power is zero.}\end{array}$

Q.6 Obtain the resonant frequency ω_r of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?

Ans.

$\begin{array}{l}\text{Here, inductance, L = 2.0 H}\\ {\text{Capacitance,C = 32 μF = 32 × 10}}^{\text{-6}}\text{F}\\ \text{Resistance, R = 10 Ω}\\ \text{Resonant frequency can\hspace{0.33em}be\hspace{0.33em}obtained by\hspace{0.33em}the\hspace{0.33em}relation,}\\ {\text{ω}}_{\text{r}}\text{=}\frac{\text{1}}{\sqrt{\text{LC}}}\\ \text{=}\frac{\text{1}}{\sqrt{{\text{2 H × 32 × 10}}^{\text{-6}}\text{F}}}\text{=}\frac{\text{1}}{{\text{8×10}}^{\text{-3}}}{\text{s}}^{\text{-1}}{\text{= 125\hspace{0.33em}s}}^{\text{-1}}\\ \text{Now, Q-value of the circuit is given by:}\\ \text{Q =}\frac{\text{1}}{\text{R}}\sqrt{\frac{\text{L}}{\text{C}}}\\ \text{=}\frac{\text{1}}{\text{10}}\sqrt{\frac{\text{2H}}{{\text{32 × 10}}^{\text{-6}}\text{F}}}\\ \text{=}\frac{\text{1}}{\text{10}}\text{×}\frac{\text{1}}{{\text{4×10}}^{\text{-3}}}\text{= 25}\\ \therefore \text{The Q-Value of this circuit is 25.}\end{array}$

Q.7 A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Ans.

$\begin{array}{l}{\text{Here,\hspace{0.33em}capacitance, C = 30 μF = 30 ×10}}^{\text{-6}}\text{\hspace{0.33em}F}\\ {\text{Inductance, L = 27 mH = 27 ×10}}^{\text{-3}}\text{H}\\ \text{Angular frequency is given by\hspace{0.33em}the\hspace{0.33em}relation,}\\ {\text{ω}}_{\text{r}}\text{=}\frac{\text{1}}{\sqrt{\text{LC}}}\text{=}\frac{\text{1}}{\sqrt{{\text{27 × 10}}^{\text{-3}}{\text{H ×30 × 10}}^{\text{-6}}\text{F}}}\\ {\text{ω}}_{\text{r}}\text{=}\frac{\text{1}}{{\text{9×10}}^{\text{-4}}}{\text{= 1.11×10}}^{\text{3}}{\text{\hspace{0.33em}rads}}^{\text{-1}}\\ \therefore \text{The angular frequency of free oscillations of the}\\ {\text{circuit is 1.11 × 10}}^{\text{3}}{\text{rads}}^{\text{-1}}\text{.}\end{array}$

Q.8 Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?

Ans.

\begin{array}{l}\text{Here,}{\text{capacitance of capacitor, C = 30 μF = 30×10}}^{\text{-6}}\text{F}\\ {\text{Inductance of inductor, L = 27 mH = 27×10}}^{\text{-3}}\text{H}\\ {\text{Charge on capacitor, Q = 6 mC = 6 × 10}}^{\text{-3}}\text{C}\\ \text{Total energy stored in the capacitor can be obtained}\text{by}\\ \text{the relation,}\\ \text{E=}\frac{\text{1}}{\text{2}}\frac{{\text{Q}}^{\text{2}}}{\text{C}}\\ \text{=}\frac{\text{1}}{\text{2}}\text{×}\frac{{\left({\text{6 × 10}}^{\text{-3}}\text{C}\right)}^{\text{2}}}{{\text{30×10}}^{\text{-6}}\text{F}}\text{=}\frac{\text{6}}{\text{10}}\text{J = 0}\text{.6}\text{J}\\ \text{At later time,}\text{total energy will remain the same}\\ \text{because energy is shared between the}\text{capacitor and}\\ \text{the inductor}\text{.}\end{array}

Q.9 A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Ans.

\begin{array}{l}\text{When the frequency of the supply power equals the}\\ \text{natural frequency of the}\text{given LCR circuit,}\text{resonance}\\ \text{takes}\text{place}\text{.}\\ \text{Here,}r\text{esistance},\text{R}=\text{2}0\Omega \\ \text{Inductance},\text{L}=\text{1}.\text{5 H}\\ \text{Capacitance},\text{C}=\text{35}\mu \text{F}=\text{35}\times \text{1}{0}^{-\text{6}}\text{F}\\ \text{AC supply voltage to the}\text{circuit,}\text{V = 200 V}\\ \text{Impedance of the circuit is given by,}\\ \text{Z =}\sqrt{{\text{R}}^{\text{2}}\text{+}{\left(\text{ωL –}\frac{\text{1}}{\text{ωC}}\right)}^{\text{2}}}\\ \text{At resonance,}\\ \text{ωL =}\frac{\text{1}}{\text{ωC}}\\ \therefore Z=R\text{=20}\text{Ω}\\ \text{Current in the circuit,}\text{I =}\frac{\text{V}}{\text{Z}}\\ \text{=}\frac{\text{200 V}}{\text{20}\Omega }\\ \text{=10}\text{A}\\ \therefore \text{The average power transferred to the circuit in one}\\ \text{complete cycle = VI}\\ \text{= 200 V × 10 A}\\ \text{= 2000 W}\end{array}

Q.10 A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?

[Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.]

Ans.

\begin{array}{l}\text{Here,}\text{lower tuning frequency,}\\ {\nu }_{\text{1}}{\text{= 800 kHz = 800 × 10}}^{\text{3}}\text{Hz}\\ \text{Upper tuning frequency,}\\ {\nu }_{\text{2}}{\text{= 1200 kHz = 1200 ×10}}^{\text{3}}\text{Hz}\\ \text{Effective inductance of LC}\text{circuit,}\\ {\text{L = 200 μH = 200 ×10}}^{\text{-6}}\text{H}\\ \text{Capacitance of the}{\text{variable capacitor for ν}}_{\text{1}}\text{is given by:}\\ {\text{C}}_{\text{1}}\text{=}\frac{\text{1}}{{\text{ω}}_{\text{1}}^{\text{2}}\text{L}}\\ {\text{Here,ω}}_{\text{1}}{\text{= Angular frequency for capacitor C}}_{\text{1}}\\ {\text{ω}}_{\text{1}}\text{= 2}\pi {\nu }_{\text{1}}\text{= 2}\pi {\text{× 800 ×10}}^{\text{3}}{\text{rads}}^{\text{-1}}\\ \therefore {\text{C}}_{\text{1}}\text{=}\frac{\text{1}}{{\left(\text{2}\pi {\text{× 800 × 10}}^{\text{3}}{\text{rads}}^{\text{-1}}\right)}^{\text{2}}{\text{× 200 ×10}}^{\text{-6}}\text{H}}\\ {\text{C}}_{\text{1}}\text{= 1}{\text{.9809×10}}^{\text{-10}}\text{F = 198}\text{.1}\text{pF}\\ {\text{Capacitance of variable capacitor for ν}}_{\text{2}}\text{is}\text{given}\text{by,}\\ {\text{C}}_{\text{2}}\text{=}\frac{\text{1}}{{\text{ω}}_{\text{2}}^{\text{2}}\text{L}}\\ {\text{Here,ω}}_{\text{2}}{\text{= Angular frequency for capacitor C}}_{\text{2}}\\ {\text{ω}}_{\text{2}}\text{= 2}\pi {\nu }_{\text{2}}\text{= 2}\pi {\text{×1200×10}}^{\text{3}}{\text{rads}}^{\text{-1}}\\ \therefore {\text{C}}_{\text{2}}\text{=}\frac{\text{1}}{{\left(\text{2}\pi {\text{×1200 ×10}}^{\text{3}}{\text{rads}}^{\text{-1}}\right)}^{\text{2}}{\text{×200×10}}^{\text{-6}}\text{H}}\\ {\text{C}}_{\text{2}}\text{= 88}\text{.04}\text{pF}\\ \therefore \text{The range of the variable capacitor is from 88}\text{.04 pF to 198}\text{.1 pF}\text{.}\end{array}

Q.11 Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L= 5.0 H, C = 80μF, R = 40 Ω
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.33em}}\mathrm{i}\text{nductance of inductor},\text{L}=\text{5}.0\text{H}\\ \text{Capacitance of capacitor},\text{C}=\text{8}0\mathrm{\mu F}\\ \mathrm{C}=\text{8}0\times \text{1}{0}^{-\text{6}}\text{F}\\ \text{Resistance of resistor},\text{R}=\text{4}0\mathrm{\Omega }\\ \text{Potential of variable voltage source},\text{V}=\text{23}0\text{V}\\ \left(\text{a}\right)\text{Resonance angular frequency is given by}:\\ {\mathrm{\omega }}_{\mathrm{R}}=\frac{1}{\sqrt{\mathrm{LC}}}\\ =\frac{1}{\sqrt{\text{5}.0\mathrm{H}\times \text{8}0\times \text{1}{0}^{-\text{6}}\mathrm{F}}}\\ =\frac{{10}^3}{20}{\mathrm{rads}}^{-1}\\ =50{\mathrm{rads}}^{-1}\\ \therefore \text{The circuit will come in resonance for a source}\\ \text{frequency of 5}0{\text{rads}}^{\text{-1}}.\\ \left(\text{b}\right)\text{Impedance of the circuit is given by},\\ \mathrm{Z}=\sqrt{{\mathrm{R}}^2+{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}^2}\\ \text{At resonance},\\ \mathrm{\omega L}=\frac{1}{\mathrm{\omega C}}\\ \therefore \mathrm{Z}=\mathrm{R}\\ =40\mathrm{\Omega }\\ \text{Amplitude of current at resonating frequency},{\mathrm{I}}_{\mathrm{o}}=\frac{{\mathrm{V}}_{\mathrm{o}}}{\mathrm{Z}}\\ \text{Here},{\text{V}}_0=\text{Peak voltage}\\ \text{=}\sqrt{2}\mathrm{V}\\ \therefore {\mathrm{I}}_{\mathrm{o}}=\frac{\sqrt{2}\mathrm{V}}{\mathrm{Z}}\\ =\frac{\sqrt{2}\times 230\mathrm{V}}{40\mathrm{\Omega }}\\ =8.13\mathrm{A}\\ \therefore \text{At resonance},\text{the impedance of the circuit is 4}0\mathrm{\Omega }\\ \text{and the amplitude of the\hspace{0.33em}current is 8}.\text{13 A}.\\ \left(\text{c}\right)\text{Across the inductor,\hspace{0.33em}}\mathrm{r}\text{ms potential drop},\\ {\left({\text{V}}_{\text{L}}\right)}_{\text{rms}}=\text{I}\times {\mathrm{\omega }}_{\text{R}}\text{L}\\ \text{Here},\text{I}=\text{rms current}\\ \text{=}\frac{{\mathrm{I}}_{\mathrm{o}}}{\sqrt{2}}\\ =\frac{\sqrt{2}\mathrm{V}}{\sqrt{2}\mathrm{Z}}\\ =\frac{230}{40}\mathrm{A}\\ {\left({\text{V}}_{\text{L}}\right)}_{\text{rms}}=\frac{230}{40}\text{A}\times 50{\mathrm{rads}}^{-1}\times 5\text{H}\\ =1437.5\mathrm{V}\\ \text{Potential drop across capacitor},\\ {\left({\text{V}}_{\mathrm{C}}\right)}_{\text{rms}}=\text{I}\times \frac{1}{{\mathrm{\omega }}_{\text{R}}\mathrm{C}}\\ =\frac{230}{40}\mathrm{A}\times \frac{1}{50{\mathrm{rads}}^{-1}\times 80\times {10}^{-6}\text{F}}\\ =1437.5\mathrm{V}\\ \text{Potential drop across\hspace{0.33em}resistor},\\ {\left({\text{V}}_{\text{R}}\right)}_{\text{rms}}=\text{IR}\\ =\frac{230}{40}\mathrm{A}\times \text{4}0\mathrm{\Omega }\\ =\text{23}0\text{V}\\ \text{Potential drop across LC combination},\\ {\mathrm{V}}_{\mathrm{LC}}=\mathrm{I}\left({\mathrm{\omega }}_{\text{R}}\text{L}-\frac{1}{{\mathrm{\omega }}_{\text{R}}\mathrm{C}}\right)\\ \text{At resonance},\\ {\mathrm{\omega }}_{\text{R}}\text{L =}\frac{1}{{\mathrm{\omega }}_{\text{R}}\mathrm{C}}\\ \therefore {\text{V}}_{\text{LC}}=0\\ \therefore \text{It is proved that the potential drop across the LC}\\ \text{combination is zero at\hspace{0.33em}resonating frequency}.\end{array}$

Q.12 An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t =0.

(a) What is the total energy stored initially? Is it conserved during LC oscillations?

(b) What is the natural frequency of the circuit?

(c) At what time is the energy stored?
(i) Completely electrical (i.e., stored in the capacitor)?
(ii) Completely magnetic (i.e., stored in the inductor)?

(d) At what times is the total energy shared equally between the inductor and the capacitor?

(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

Ans.

$\begin{array}{l}{\text{Here,\hspace{0.33em}inductance\hspace{0.33em}of\hspace{0.33em}inductor,L = 20\hspace{0.33em}mH = 20×10}}^{\text{-3}}\text{\hspace{0.33em}H}\\ {\text{Capacitance\hspace{0.33em}of\hspace{0.33em}capacitor,\hspace{0.33em}C = 50\hspace{0.33em}μF = 50 × 10}}^{\text{-6}}\text{\hspace{0.33em}F}\\ {\text{Initial\hspace{0.33em}charge\hspace{0.33em}on\hspace{0.33em}capacitor,Q = 10\hspace{0.33em}mC = 10 × 10}}^{\text{-3}}\text{\hspace{0.33em}C}\\ \text{(a)\hspace{0.33em}Total\hspace{0.33em}energy\hspace{0.33em}stored\hspace{0.33em}initially\hspace{0.33em}in\hspace{0.33em}the\hspace{0.33em}circuit\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}by the\hspace{0.33em}relation,}\\ \text{E =}\frac{\text{1}}{\text{2}}\frac{{\text{Q}}^{\text{2}}}{\text{C}}\\ \text{=}\frac{{\left({\text{10×10}}^{\text{-3}}\text{C}\right)}^{\text{2}}}{{\text{2×50×10}}^{\text{-6}}\text{F}}\text{= 1\hspace{0.33em}J}\\ \text{Since\hspace{0.33em}there\hspace{0.33em}is\hspace{0.33em}no\hspace{0.33em}resistor\hspace{0.33em}connected\hspace{0.33em}in\hspace{0.33em}the\hspace{0.33em}given LC\hspace{0.33em}circuit, therefore total\hspace{0.33em}}\\ \text{energy stored\hspace{0.33em}initially\hspace{0.33em}in\hspace{0.33em}the\hspace{0.33em}given\hspace{0.33em}LC\hspace{0.33em}circuit\hspace{0.33em}is conserved\hspace{0.33em}during\hspace{0.33em}oscillations.}\\ \text{(b)\hspace{0.33em}The\hspace{0.33em}relation\hspace{0.33em}for\hspace{0.33em}natural\hspace{0.33em}frequency\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}circuit\hspace{0.33em}is given\hspace{0.33em}as:}\\ \mathrm{\nu }\text{=}\frac{\text{1}}{\text{2}\mathrm{\pi }\sqrt{\text{LC}}}\\ \mathrm{\nu }\text{=}\frac{\text{1}}{\text{2}\mathrm{\pi }\sqrt{{\text{20×10}}^{\text{-3}}{\text{H × 50×10}}^{\text{-6}}\mathrm{F}}}\\ \mathrm{\nu }\text{=}\frac{{\text{10}}^{\text{3}}}{\text{2}\mathrm{\pi }}\text{= 159.24\hspace{0.33em}Hz}\\ \text{Natural\hspace{0.33em}angular\hspace{0.33em}frequency\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:}\\ {\text{ω}}_{\text{r}}\text{=}\frac{\text{1}}{\sqrt{\text{LC}}}\\ \text{=}\frac{\text{1}}{\sqrt{{\text{20×10}}^{\text{-3}}{\text{H× 50 ×10}}^{\text{-6}}}\mathrm{F}}\text{=}\frac{\text{1}}{\sqrt{{\text{10}}^{\text{-6}}}}\\ {\text{=10}}^{\text{3}}{\text{\hspace{0.33em}rads}}^{\text{-1}}\\ \therefore {\text{Natural\hspace{0.33em}angular\hspace{0.33em}frequency\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}given\hspace{0.33em}circuit\hspace{0.33em}is 10}}^{\text{3}}{\text{\hspace{0.33em}rads}}^{\text{-1}}.\\ \text{(c)\hspace{0.33em}(i)\hspace{0.33em}At\hspace{0.33em}any\hspace{0.33em}instant\hspace{0.33em}t,total\hspace{0.33em}charge\hspace{0.33em}on\hspace{0.33em}the\hspace{0.33em}capacitor\hspace{0.33em}is given\hspace{0.33em}as:}\\ \text{Q’ = Q cos}\frac{\text{2}\mathrm{\pi }}{\text{T}}\text{t}\\ \text{For\hspace{0.33em}electrical\hspace{0.33em}energy\hspace{0.33em}stored,we\hspace{0.33em}can\hspace{0.33em}write,}\\ \text{Q’ = Q}\\ \therefore \text{It can be concluded that the energy stored in the}\\ \text{capacitor is completely electrical\hspace{0.33em}at time,}\\ \text{t = 0,}\frac{\text{T}}{\text{2}}\text{,T,}\frac{\text{3T}}{\text{2}}\text{,}.\dots ..\\ \left(\text{ii}\right)\text{Magnetic energy is the maximum when electrical energy,Q’ = 0.}\\ \therefore \text{It can be concluded that the energy stored in the}\\ \text{capacitor is completely magnetic at time,}\\ \text{t =}\frac{\text{T}}{\text{4}}\text{,}\frac{\text{3T}}{\text{4}}\text{,}\frac{\text{5T}}{\text{4}}\text{,}.\dots ..\\ \left(\text{d}\right){\text{Q}}^{\text{1}}\text{= Charge on capacitor when total energy is\hspace{0.33em}equally shared between}\\ \text{the capacitor and the inductor at time t.}\\ \text{When total energy is equally shared between the inductor and capacitor,}\\ \text{Energy\hspace{0.33em}stored in the capacitor=}\frac{\text{1}}{\text{2}}\left(\text{maximum energy}\right)\\ \Rightarrow \frac{\text{1}}{\text{2}}\frac{{\left({\text{Q}}^{\text{1}}\right)}^{\text{2}}}{\text{C}}\text{=}\frac{\text{1}}{\text{2}}\left(\frac{\text{1}}{\text{2}}\frac{{\text{Q}}^{\text{2}}}{\text{C}}\right)\text{=}\frac{\text{1}}{\text{4}}\frac{{\text{Q}}^{\text{2}}}{\text{C}}\\ \left({\text{Here,\hspace{0.33em}Q}}^{\text{1}}\text{=}\frac{\text{Q}}{\sqrt{\text{2}}}\right)\\ {\text{From\hspace{0.33em}Q}}^{\text{1}}\text{= Q cos}\frac{\text{2}\mathrm{\pi }}{\text{T}}\text{t,\hspace{0.33em}we\hspace{0.33em}have}\\ \frac{\text{Q}}{\sqrt{\text{2}}}\text{= Q cos}\frac{\text{2}\mathrm{\pi }}{\text{T}}\text{t}\\ \cos \frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{t}=\frac{1}{\sqrt{2}}=\cos \left(2\mathrm{n}+1\right)\frac{\mathrm{\pi }}{4}\\ \mathrm{Where}\mathrm{n}=0,1,2,3,...\\ \therefore \mathrm{t}=\left(2\mathrm{n}+1\right)\frac{\mathrm{T}}{8}\\ \therefore \text{Total energy is equally shared between the inductor and the capacity at time},\\ \mathrm{t}=\frac{\mathrm{T}}{8},\frac{3\mathrm{T}}{8},\frac{5\mathrm{T}}{8},.\dots ..\\ \left(\text{e}\right)\text{The\hspace{0.33em}presence\hspace{0.33em}of\hspace{0.33em}a\hspace{0.33em}resistor in the circuit\hspace{0.33em}involves},\text{dissipation\hspace{0.33em}of\hspace{0.33em}energy as}\\ \text{heat energy in the circuit}.\text{The resistance damps out the LC oscillations\hspace{0.33em}and\hspace{0.33em}}\\ \text{they\hspace{0.33em}disappear\hspace{0.33em}when\hspace{0.33em}total\hspace{0.33em}energy\hspace{0.33em}of\hspace{0.33em}1\hspace{0.33em}J\hspace{0.33em}is\hspace{0.33em}dissipated\hspace{0.33em}in\hspace{0.33em}the\hspace{0.33em}form\hspace{0.33em}heat.}\end{array}$

Q.13 A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply.

(a) What is the maximum current in the coil?

(b) What is the time lag between the voltage maximum and the current maximum?

Ans.

\begin{array}{l}\text{Inductance of inductor, L = 0}\text{.50 H}\\ \text{Resistance of resistor, R = 100 Ω}\\ \text{Potential of supply voltage, V = 240 V}\\ \text{Frequency of the}\text{supply}\text{voltage,}\nu \text{= 50 Hz}\\ \left(\text{a}\right)\text{Peak voltage is given by}\text{the}\text{relation,}\\ {\text{V}}_{\text{0}}\text{=}\sqrt{\text{2}}\text{V =}\sqrt{\text{2}}\text{× 240 V = 339}\text{.41}\text{V}\\ \text{Angular frequency of the supply}\text{voltage}\text{is}\text{given}\text{as:}\\ \text{ω = 2}\pi \nu \text{= 2}\pi \text{× 50 Hz}\\ \text{= 100}\pi {\text{rads}}^{\text{-1}}\\ \text{Maximum current in the circuit,}{\text{I}}_{\text{0}}\text{=}\frac{{\text{V}}_{\text{0}}}{\sqrt{{\text{R}}^{\text{2}}{\text{+ ω}}^{\text{2}}{\text{L}}^{\text{2}}}}\\ \text{=}\frac{\text{339}\text{.41 V}}{\sqrt{{\left(\text{100 Ω}\right)}^{\text{2}}\text{+}{\left(\text{100}\pi {\text{rads}}^{\text{-1}}\right)}^{\text{2}}{\left(\text{0}\text{.50 H}\right)}^{\text{2}}}}\\ \text{=1}\text{.82}\text{A}\\ \left(\text{b}\right)\text{Voltage is given by}\text{the}\text{relation,}\\ {\text{V = V}}_{\text{0}}\text{cos ωt}\\ \text{Current is given by}\text{the}\text{relation,}\\ {\text{I = I}}_{\text{0}}\text{cos}\left(\text{ωt –}\varphi \right)\\ \text{Here,}\varphi \text{= Phase difference between voltage and current}\\ \text{At time, t = 0}\\ {\text{V= V}}_{\text{0}}\left(\text{voltage is maximum}\right)\\ \text{For}\text{ωt –}\varphi \text{=0 i}\text{.e}\text{.,at time t =}\frac{\varphi }{\text{ω}}{\text{, I = I}}_{\text{0}}\left(\text{current is maximum}\right)\\ \therefore \text{The time lag between maximum voltage and maximum current =}\frac{\varphi }{\text{ω}}\\ \text{The}\text{relation}\text{for}\text{the}\text{phase angle}\varphi \text{is given as:}\\ \text{tan}\varphi \text{=}\frac{\text{ωL}}{\text{R}}\text{=}\frac{\text{2}\pi \text{×50 Hz ×0}\text{.5 H}}{\text{100 Ω}}\text{=1}\text{.57}\\ \therefore \varphi \text{= 57}\text{.5°=}\frac{\text{57}\text{.5}\varphi }{\text{180}}\text{rad}\\ \therefore \text{ωt =}\frac{\text{57}\text{.5}\pi }{\text{180}}\\ \therefore \text{t =}\frac{\text{57}\text{.5}\pi }{\text{180 × 2}\pi \text{×50 Hz}}\\ \text{= 3}{\text{.19 ×10}}^{\text{-3}}\text{s = 3}\text{.2}\text{ms}\\ \therefore \text{The time lag between maximum voltage and maximum current is 3}\text{.2 ms}\text{.}\end{array}

Q.14 Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

Ans.

\begin{array}{l}\text{Here,}\text{inductance of inductor, L = 0}\text{.50 H}\\ \text{Resistance of resistor, R = 100 Ω}\\ \text{Potential of supply voltages, V = 240 V}\\ \text{Frequency of supply},\nu =\text{1}0\text{kHz}\\ =\text{1}{0}^4\text{Hz}\\ \text{Angular frequency}\text{of}\text{supply},\omega =\text{2}\pi \nu \\ =\text{2}\pi \times \text{1}{0}^{\text{4}}{\text{rads}}^{\text{-1}}\\ \left(\text{a}\right)\text{Peak voltage}\text{is}\text{given}\text{by}\text{the}\text{relation},\\ V_0=\sqrt{2}\times V\\ =240\sqrt{2}V\\ \text{Maximum current}\text{is}\text{given}\text{by},I_0=\frac{V_0}{\sqrt{R^2+{\omega }^2{L}^2}}\\ =\frac{240\sqrt{2}\text{V}}{\sqrt{{\left(100\Omega \right)}^2+{\left(2\pi \times {10}^4{\text{rads}}^{\text{-1}}\right)}^2\times {\left(0.50H\right)}^2}}\\ =1.1\times {10}^{-2}A\\ \left(\text{b}\right)\text{The}\text{relation}f\text{or phase difference}\Phi isgivenas:\\ \tan \varphi =\frac{\omega L}{R}\\ =\frac{2\pi \times {10}^4{\text{rads}}^{\text{-1}}\times 0.50\text{H}}{100\Omega }\\ =100\pi \\ \varphi ={89.82}^o\\ =\frac{89.82\pi }{180}rad\\ \omega t=\frac{89.82\pi }{180}\text{rad}\\ \therefore t=\frac{89.82\pi \text{rad}}{180\times 2\pi \times {10}^4{\text{rads}}^{\text{-1}}}\\ =25\mu s\\ \text{We}{\text{observe that, I}}_0\text{is very small in this case}.\\ \therefore \text{At high frequencies},\text{the}\text{inductor amounts to}\\ \text{an open circuit}.\\ \text{In a dc circuit},\text{after a steady state}\text{is}\text{reached},\\ \omega =0.\\ \therefore {\text{X}}_{\text{L}}\text{=}\omega \text{L}\\ \text{= 0}\\ \therefore \text{Inductor L behaves like a}\text{pure conducting object}.\end{array}

Q.15 A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.

(a) What is the maximum current in the circuit?

(b) What is the time lag between the current maximum and the voltage maximum?

Ans.

\begin{array}{l}\text{Here,}{\text{capacitance of capacitor, C = 100 μF = 100 ×10}}^{\text{-6}}\text{F}\\ \text{Resistance of resistor, R = 40 Ω}\\ \text{Supply voltage, V = 110 V}\\ \left(\text{a}\right)\text{Here,}\text{frequency of oscillations,}\nu \text{= 60 Hz}\\ \text{Angular frequency,ω = 2π}\nu \text{= 2}\pi \text{× 60}{\text{rads}}^{\text{-1}}\\ \text{In a RC circuit, the relation for impedance is}\text{given}\text{as:}\\ \text{Z=}\sqrt{{\text{R}}^{\text{2}}\text{+}\frac{\text{1}}{{\text{ω}}^{\text{2}}{\text{C}}^{\text{2}}}}\\ \text{Peak voltage}\text{is}\text{given}{\text{as, V}}_{\text{0}}\text{= V}\sqrt{\text{2}}\text{= 110}\sqrt{\text{2}}\text{V}\\ \text{Maximum current is given by,}{\text{I}}_{\text{0}}\text{=}\frac{{\text{V}}_{\text{0}}}{\text{Z}}\\ \text{=}\frac{{\text{V}}_{\text{0}}}{\sqrt{{\text{R}}^{\text{2}}\text{+}\frac{\text{1}}{{\text{ω}}^{\text{2}}{\text{C}}^{\text{2}}}}}\\ \text{=}\frac{\text{110}\sqrt{\text{2}}V}{\sqrt{{\left(\text{40}\Omega \right)}^{\text{2}}\text{+}\frac{\text{1}}{{\left(\text{120}\pi {\text{rads}}^{\text{-1}}\right)}^{\text{2}}{\left({\text{10}}^{\text{-4}}F\right)}^{\text{2}}}}}\\ \text{=}\frac{\text{110}\sqrt{\text{2}}\text{V}}{\sqrt{\text{1600}{\Omega }^2\text{+}\frac{{\text{10}}^{\text{8}}}{{\left(\text{120}\pi \right)}^{\text{2}}}{\Omega }^2}}\text{= 3}\text{.24}\text{A}\\ \left(\text{b}\right)\text{In the RC circuit, the voltage lags behind the}\\ \text{current by}\text{phase angle}\varphi \text{.}\\ \text{Phase}\text{angle is given by the relation:}\\ \text{tan}\varphi \text{=}\frac{\frac{\text{1}}{\text{ωC}}}{\text{R}}\text{=}\frac{\text{1}}{\text{ωCR}}\\ \text{=}\frac{\text{1}}{\text{120}\pi {\text{rads}}^{\text{-1}}{\text{×10}}^{\text{-4}}\text{F× 40}\Omega }\text{= 0}\text{.6635}\\ \therefore \varphi {\text{= tan}}^{\text{-1}}\left(\text{0}\text{.6635}\right)\text{= 33}{\text{.56}}^{\text{o}}\\ \text{=}\frac{\text{33}\text{.56}\pi }{\text{180}}\text{rad}\\ \therefore \text{Time}\text{lag =}\frac{\varphi }{\text{ω}}\\ \text{=}\frac{\text{33}\text{.56}\pi }{\text{180 × 120}\pi }\text{= 1}{\text{.55 × 10}}^{\text{-3}}\text{s = 1}\text{.55}\text{ms}\\ \therefore \text{The}\text{time lag between maximum current and maximum voltage is 1}\text{.55 ms}\text{.}\end{array}

Q.16 Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.

Ans.

\begin{array}{l}\text{Here,}\text{capacitance of capacitor,}\\ {\text{C = 100 μF = 100×10}}^{\text{-6}}\text{F}\\ \text{Resistance of resistor, R = 40 Ω}\\ \text{Supply voltage,V = 110 V}\\ \text{Frequency of supply}\text{voltage,}\nu {\text{= 12 kHz =12×10}}^{\text{3}}\text{Hz}\\ \text{Angular Frequency, ω = 2}\pi \nu \text{= 2 ×}\pi {\text{× 12 ×10}}^{\text{3}}{\text{rads}}^{\text{-1}}\\ \text{= 24}\pi {\text{×10}}^{\text{3}}{\text{rads}}^{\text{-1}}\\ \text{Peak voltage}\text{is}\text{given}\text{by}\text{the}\text{relation,}{\text{V}}_{\text{0}}\text{= V}\sqrt{\text{2}}\text{= 110}\sqrt{\text{2}}\\ \text{Maximum current}\text{is}\text{given}\text{by,}\\ {\text{I}}_{\text{0}}\text{=}\frac{{\text{V}}_{\text{0}}}{\sqrt{{\text{R}}^{\text{2}}\text{+}\frac{\text{1}}{{\text{ω}}^{\text{2}}{\text{C}}^{\text{2}}}}}\\ \text{=}\frac{\text{110}\sqrt{\text{2}}}{\sqrt{{\left(\text{40}\Omega \right)}^{\text{2}}\text{+}\frac{\text{1}}{{\left(\text{24}\pi {\text{×10}}^{\text{3}}{\text{rads}}^{\text{-1}}\right)}^{\text{2}}{\left({\text{100×10}}^{\text{-6}}\text{F}\right)}^{\text{2}}}}}\\ \text{=}\frac{\text{110}\sqrt{\text{2}}}{\sqrt{\text{1600+}{\left(\frac{\text{10}}{\text{24}\pi }\right)}^{\text{2}}}}\text{= 3}\text{.9}\text{A}\\ \text{In an RC circuit, the voltage lags behind the current}\\ \text{by phase angle}\varphi \text{.}\\ \text{The}\text{relation}\text{for}\text{phase angle}\varphi \text{is}\text{given as:}\\ \text{tan}\varphi \text{}\text{}\text{=}\frac{\frac{\text{1}}{\text{ωC}}}{\text{R}}\text{=}\frac{\text{1}}{\text{ωCR}}\\ \text{=}\frac{\text{1}}{\text{24}\pi {\text{×10}}^{\text{3}}{\text{rads}}^{\text{-1}}{\text{× 100 ×10}}^{\text{-6}}\text{F × 40}\Omega }\\ \text{tan}\varphi \text{=}\frac{\text{1}}{\text{96}\pi }\\ \therefore \varphi \simeq \text{0}\text{.2}°\text{=}\frac{\text{0}\text{.2}\pi }{\text{180}}\text{rad}\\ \therefore \text{Time}\text{lag =}\frac{\varphi }{\text{ω}}\\ \text{=}\frac{\text{0}\text{.2}\pi }{\text{180× 24}\pi {\text{×10}}^{\text{3}}{\text{rads}}^{\text{-1}}}\\ \text{= 1}{\text{.55×10}}^{\text{-3}}\text{s = 0}\text{.04}\text{μs}\\ \therefore \varphi \text{tends to become zero at high frequencies}\text{.}\\ \therefore \text{At very high frequencies, capacitor C}\text{acts as a}\\ \text{conductor}\text{of}\text{negligible}\text{capacitive}\text{reactance}\text{.}\\ \text{In a}\text{dc circuit, after the steady state, ω = 0}\text{.}\\ \therefore {\text{X}}_{\text{C}}\text{=}\frac{\text{1}}{\text{ωC}}\to \infty \\ \therefore \text{Capacitor C amounts to}\text{an open circuit}\text{.}\end{array}

Q.17 Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.

Ans.

$\begin{array}{l}\text{An inductor}\left(\text{L}\right),\text{a capacitor}\left(\text{C}\right),\text{and a resistor}\left(\text{R}\right)\text{is}\\ \text{connected in parallel with each other in a circuit where},\\ \text{Inductance\hspace{0.33em}of\hspace{0.33em}inductor\hspace{0.33em}L}=\text{5}.0\text{H}\\ \text{Capacitance\hspace{0.33em}of\hspace{0.33em}capacitor\hspace{0.33em}C}=\text{8}0\mathrm{\mu }\text{F}\\ =\text{8}0\times \text{1}{0}^{-\text{6}}\text{F}\\ \text{Resistance\hspace{0.33em}of\hspace{0.33em}resistor\hspace{0.33em}R}=\text{4}0\mathrm{\Omega }\\ \text{Potential of voltage source},\\ \text{V}=\text{23}0\text{V}\\ \text{The\hspace{0.33em}relation\hspace{0.33em}for\hspace{0.33em}}\mathrm{i}\text{mpedance}\left(\text{Z}\right)\text{of the given parallel}\\ \text{LCR circuit is given as}:\\ \frac{1}{\mathrm{Z}}=\sqrt{\frac{1}{{\mathrm{R}}^2}+{\left(\frac{1}{\mathrm{\omega L}}-\mathrm{\omega C}\right)}^2}\\ \text{Here},\\ \mathrm{\omega }=\text{Angular frequency}\\ \text{At resonance},\\ \frac{1}{\mathrm{\omega L}}-\mathrm{\omega C}=0\\ \therefore \mathrm{\omega }=\frac{1}{\sqrt{\mathrm{LC}}}\\ =\frac{1}{\sqrt{5\text{H}\times 80\times {10}^{-6}\mathrm{F}}}\\ =50{\mathrm{rads}}^{-1}\\ \therefore \text{Z is the maximum at 5}0{\text{rads}}^{\text{-1}}.\\ \therefore \mathrm{T}\text{otal current is minimum}.\\ \text{Rms current flowing through inductor L is given by,}\\ {\text{I}}_{\text{L}}\text{=}\frac{\mathrm{V}}{\mathrm{\omega L}}\\ =\frac{230\mathrm{V}}{50{\text{rads}}^{\text{-1}}\times 5\text{H}}\\ =0.92\mathrm{A}\\ \text{Rms current flowing through capacitor C is given by,}\\ {\text{I}}_{\text{C}}=\frac{\mathrm{V}}{\frac{1}{\mathrm{\omega C}}}\\ =\mathrm{\omega CV}\\ =50{\text{rads}}^{\text{-1}}\times 80\times {10}^{-6}\text{F}\times 230\text{V}\\ =0.92\mathrm{A}\\ \text{Rms current flowing through resistor R is given by,}\\ {\text{I}}_{\mathrm{R}}=\frac{\mathrm{V}}{\mathrm{R}}\\ =\frac{230\mathrm{V}}{40\mathrm{\Omega }}\\ =5.75\mathrm{A}\end{array}$

Q.18 A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.
(a) Obtain the current amplitude and rms values.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]

Ans.

$\begin{array}{l}\text{Here,\hspace{0.33em}Inductance},\text{L}=\text{8}0\text{mH}\\ =\text{8}0\times \text{1}{0}^{-\text{3}}\text{H}\\ \text{Capacitance},\text{C}=\text{6}0\mathrm{\mu }\text{F}\\ =\text{6}0\times \text{1}{0}^{-\text{6}}\text{F}\\ \text{Supply voltage},\text{V}=\text{23}0\text{V}\\ \text{Frequency},\mathrm{\nu }=\text{5}0\text{Hz}\\ \text{Angular frequency},\mathrm{\omega }=\text{2}\mathrm{\pi \nu }\\ =\text{1}00\mathrm{\pi }{\text{rads}}^{\text{-1}}\\ \text{Peak voltage\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}by},{\text{V}}_0=\text{V}\sqrt{2}\\ =\text{23}0\sqrt{2}\text{\hspace{0.33em}V}\\ \left(\text{a}\right)\text{Maximum current is given by\hspace{0.33em}the\hspace{0.33em}relation,}\\ {\text{I}}_{\text{0}}=\frac{{\mathrm{V}}_0}{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}\\ =\frac{\text{23}0\sqrt{2}}{\left(\text{1}00\mathrm{\pi }{\text{rads}}^{\text{-1}}\times \text{8}0\times \text{1}{0}^{-\text{3}}\mathrm{H}-\frac{1}{\text{1}00\mathrm{\pi }\times \text{6}0\times \text{1}{0}^{-\text{6}}\mathrm{L}}\right)}\\ =\frac{\text{23}0\sqrt{2}\text{V}}{\left(8\mathrm{\pi }-\frac{1000}{6\mathrm{\pi }}\right)\mathrm{\Omega }}\\ =-11.63\mathrm{A}\\ \text{Negative sign appears because\hspace{0.33em}}\mathrm{\omega L}<\frac{1}{\mathrm{\omega C}}\\ \text{Amplitude of the\hspace{0.33em}maximum current},\left|{\text{I}}_{\text{0}}\right|=11.63\mathrm{A}\\ \therefore \text{Rms value of current},\mathrm{I}=\frac{{\text{I}}_{\text{0}}}{\sqrt{2}}\\ =\frac{-11.63\text{A}}{\sqrt{2}}\\ =-8.22\mathrm{A}\\ \left(\text{b}\right)\text{Potential difference across the inductor\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}by},{\text{\hspace{0.17em}\hspace{0.17em}V}}_{\text{L}}=\text{I}\times \mathrm{\omega }\text{L}\\ =\text{8}.\text{22 A}\times \text{1}00\mathrm{\pi }{\text{rads}}^{\text{-1}}\times \text{8}0\times \text{1}{0}^{-\text{3}}\text{H}\\ =\text{2}0\text{6}.\text{61 V}\\ \text{Potential difference across the capacitor\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}by},\\ {\text{V}}_{\mathrm{C}}=\text{I}\times \frac{1}{\mathrm{\omega C}}\\ =8.22\times \frac{1}{\text{1}00\mathrm{\pi }{\text{rads}}^{\text{-1}}\times \text{6}0\times \text{1}{0}^{-6}\text{F}}\\ =436.3\mathrm{V}\\ \left(\text{c}\right)\text{Average power transferred over\hspace{0.33em}a\hspace{0.33em}complete\hspace{0.33em}cycle\hspace{0.33em}}\\ \text{by the source\hspace{0.33em}to\hspace{0.33em}inductor is zero as\hspace{0.33em}actual voltage}\\ \text{leads the current\hspace{0.33em}by}\frac{\mathrm{\pi }}{2}\mathrm{in}\mathrm{the}\text{\hspace{0.33em}inductor\hspace{0.33em}}.\\ \left(\text{d}\right)\text{Average power transferred over\hspace{0.33em}a\hspace{0.33em}complete\hspace{0.33em}cycle\hspace{0.33em}}\\ \text{by the source\hspace{0.33em}to\hspace{0.33em}the capacitor is zero as voltage lags}\\ \text{current by}\frac{\mathrm{\pi }}{2}\mathrm{in}\mathrm{the}\mathrm{capacitor}.\\ \left(\text{e}\right)\text{The\hspace{0.33em}}\mathrm{t}\text{otal average\hspace{0.33em}power absorbed by\hspace{0.33em}the\hspace{0.33em}circuit\hspace{0.33em}is}\\ \text{zero}.\end{array}$

Q.19 Suppose the circuit in Exercise 7.18 has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Ans.

\begin{array}{l}\text{Here,}\text{inductance of the}\text{inductor,}\\ \text{L = 80 mH}\\ {\text{= 80 × 10}}^{\text{-3}}\text{H}\\ \text{Capacitance of the}\text{capacitor,}\\ \text{C = 60 μF}\\ {\text{=60×10}}^{\text{-6}}\text{F}\\ \text{Resistance of the}\text{resistor,}\\ \text{R = 15 Ω}\\ \text{Potential of the}\text{voltage supply,}\\ \text{V = 230 V}\\ \text{Frequency of the}\text{signal,}\\ \text{ν = 50 Hz}\\ \text{Angular frequency of the}\text{signal,}\\ \text{ω = 2}\pi \text{ν}\\ =\text{2}\pi \times \left(\text{5}0\text{Hz}\right)\\ =\text{1}00\pi {\text{rads}}^{\text{-1}}\\ \\ \text{Average power transferred to}resistor=\text{788}.\text{44 W}\\ \text{Average power transferred to capacitor}=0\text{W}\\ \text{Total power absorbed by circuit}=\text{788}.\text{44 W}\\ \text{As}\text{elements are connected in series to each other,}\\ \therefore \text{Impedance of the circuit is}\text{given as}:\\ \\ Z=\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}\\ =\sqrt{{\left(15\right)}^2+{\left(100\pi \left(\text{8}0\times \text{1}{0}^{-\text{3}}H\right)-\frac{1}{\left(\text{1}00\pi \times \text{6}0\times \text{1}{0}^{-\text{6}}\text{F}\right)}\right)}^2}\\ =\sqrt{{\left(15\Omega \right)}^2+{\left(25.12\Omega -53.08\Omega \right)}^2}\\ =31.728\Omega \\ \text{Current flowing in the circuit}\text{is}\text{given}\text{by},\\ I=\frac{V}{Z}\\ =\frac{230V}{31.728\Omega }\\ =7.25A\\ \text{Average power transferred to resistance is given by}\\ \text{the}\text{relation,}\\ {\text{P}}_{\text{R}}={\text{I}}^{\text{2}}\text{R}\\ ={\left(\text{7}.\text{25 A}\right)}^{\text{2}}\times \text{15}\Omega \\ =\text{788}.\text{44 W}\\ \text{Average power transferred to capacitor},\\ {\text{P}}_{\text{C}}=0\\ \text{Average power transferred to inductor},\\ {\text{P}}_{\text{L}}=0\\ \text{Total power absorbed per}\text{cycle}\text{by the circuit}:\\ ={\text{P}}_{\text{R}}+{\text{P}}_{\text{C}}+{\text{P}}_{\text{L}}\\ =\text{788}.\text{44}+0+0\\ =\text{788}.\text{44 W}\\ \text{Thus, the total power absorbed by the circuit is 788}.\text{44 W}.\end{array}

Q.20 A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
(d) What is the Q-factor of the given circuit?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.33em}Inductance},\text{L}=0.\text{12 H}\\ \text{Capacitance},\text{C}=\text{48}0\text{nF}=\text{48}0\times \text{1}{0}^{-\text{9}}\text{F}\\ \text{Resistance},\text{R}=\text{23}\mathrm{\Omega }\\ \text{Supply voltage},\text{V}=\text{23}0\text{V}\\ \text{Peak voltage is given\hspace{0.33em}by,\hspace{0.33em}}\\ {\text{V}}_0=\sqrt{2}\text{V =}\sqrt{2}\times \text{23}0\text{V}\\ =\text{325}.\text{22 V}\\ \left(\text{a}\right)\text{Current flowing in the circuit is given by},\\ {\mathrm{I}}_0=\frac{{\mathrm{V}}_0}{\sqrt{{\mathrm{R}}^2+{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}^2}}\\ \text{Here},{\text{I}}_0=\text{maximum at resonance}\\ \text{At resonance},\\ {\mathrm{\omega }}_{\mathrm{R}}\mathrm{L}-\frac{1}{{\mathrm{\omega }}_{\mathrm{R}}\mathrm{C}}=0\\ \text{Here},{\mathrm{\omega }}_{\text{R}}=\text{Resonance angular frequency}\\ \therefore {\mathrm{\omega }}_{\mathrm{R}}=\frac{1}{\sqrt{\mathrm{LC}}}\\ =\frac{1}{\sqrt{0.12\mathrm{H}\times 480\times {10}^{-9}\mathrm{F}}}\\ =4166.67{\mathrm{rads}}^{-1}\\ \therefore \text{Resonant frequency\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}by\hspace{0.33em}the\hspace{0.33em}relation,}\\ {\mathrm{\nu }}_{\text{R}}\text{=}\frac{{\mathrm{\omega }}_{\mathrm{R}}}{2\mathrm{\pi }}\\ =\frac{4166.67{\mathrm{rads}}^{-1}}{2\times 3.14}\\ =663.48\mathrm{Hz}\\ \text{Maximum current\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}by\hspace{0.33em}the\hspace{0.33em}relation,}\\ {\left({\mathrm{I}}_0\right)}_{\max }=\frac{{\mathrm{V}}_0}{\mathrm{R}}\\ =\frac{325.22\mathrm{V}}{23\mathrm{\Omega }}\\ =14.14\mathrm{A}\\ \left(\text{b}\right)\text{Maximum average power absorbed by the circuit}\\ \text{is given by\hspace{0.33em}the\hspace{0.33em}relation,}\\ {\left({\mathrm{P}}_{\mathrm{av}}\right)}_{\max }=\frac{1}{2}{{\left({\mathrm{I}}_0\right)}^2}_{\max }\mathrm{R}\\ =\frac{1}{2}\times {\left(14.14\mathrm{A}\right)}^2\times 23\mathrm{\Omega }\\ =2299.3\mathrm{W}\\ \therefore \text{Resonant frequency}\left({\mathrm{\nu }}_{\text{R}}\right)\text{is\hspace{0.33em}}663.48\mathrm{Hz}.\\ \left(\text{c}\right)\text{Power transferred to the circuit is half at resonant}\\ \text{frequency,\hspace{0.33em}when}\\ \mathrm{\Delta \omega }\text{=}\frac{\mathrm{R}}{2\mathrm{L}}\\ =\frac{23\mathrm{\Omega }}{2\times 0.12\text{H}}\\ =95.83{\mathrm{rads}}^{-1}\\ \therefore \text{Frequencies at which power transferred is half}\\ \text{=}{\mathrm{\omega }}_{\text{r}}\pm \mathrm{\Delta \omega }\\ =4166.7{\mathrm{rads}}^{-1}\pm 95.83{\mathrm{rads}}^{-1}\\ =4262.53{\mathrm{rads}}^{-1}\mathrm{and}4070.87{\mathrm{rads}}^{-1}\\ \mathrm{At}\mathrm{these}\mathrm{frequencies},\mathrm{current}\mathrm{amplitude}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{I}‘=\frac{1}{\sqrt{2}}\times {\left({\mathrm{I}}_0\right)}_{\max }\\ =\frac{14.14\mathrm{A}}{\sqrt{2}}\\ =10\mathrm{A}\\ \left(\text{d}\right)\text{Q}-\text{factor of the given circuit is\hspace{0.33em}given\hspace{0.33em}by\hspace{0.33em}the\hspace{0.33em}}\\ \text{relation},\mathrm{Q}=\frac{{\mathrm{\omega }}_{\mathrm{R}}\mathrm{L}}{\mathrm{R}}\\ =\frac{4166.7{\mathrm{rads}}^{-1}\times 0.12\text{H}}{23\mathrm{\Omega }}\\ =\text{21}.\text{74}\\ \therefore \text{The Q}-\text{factor of the given circuit is 21}.\text{74}.\end{array}$

Q.21 Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 μF, and R = 7.4 Ω. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.

Ans.

\begin{array}{l}\text{Here, inductance, L = 2}\text{.0 H}\\ {\text{Capacitance,C = 32 μF = 32 × 10}}^{\text{-6}}\text{F}\\ \text{Resistance, R = 10 Ω}\\ \text{Resonant frequency can}\text{be}\text{obtained by}\text{the}\text{relation,}\\ {\text{ω}}_{\text{r}}\text{=}\frac{\text{1}}{\sqrt{\text{LC}}}\\ \text{=}\frac{\text{1}}{\sqrt{{\text{2 H × 32 × 10}}^{\text{-6}}\text{F}}}\text{=}\frac{\text{1}}{{\text{8×10}}^{\text{-3}}}{\text{s}}^{\text{-1}}\text{= 125}{\text{s}}^{\text{-1}}\\ \text{Now, Q-value of the circuit is given by:}\\ \text{Q =}\frac{\text{1}}{\text{R}}\sqrt{\frac{\text{L}}{\text{C}}}\\ \text{Q =}\frac{\text{1}}{\text{10}}\sqrt{\frac{\text{2H}}{{\text{32 × 10}}^{\text{-6}}\text{F}}}\text{=}\frac{\text{1}}{\text{10}}\text{×}\frac{\text{1}}{{\text{4×10}}^{\text{-3}}}\text{= 25}\\ \therefore \text{The Q-Value of this circuit is 25}\text{.}\end{array}

Q.22 A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?

Ans.

\begin{array}{l}\text{Here,}{\text{Input voltage, V}}_{\text{i}}\text{= 2300 V}\\ {\text{Number of turns in primary coil, n}}_{\text{p}}\text{= 4000}\\ {\text{Output voltage, V}}_{\text{o}}\text{= 230 V}\\ \text{Let}{\text{number of turns in secondary coil = n}}_{\text{s}}\\ \text{The}\text{relation}\text{between}\text{voltage}\text{and}\text{number of turns is}\\ \text{given}\text{as:}\\ \frac{{\text{V}}_{\text{i}}}{{\text{V}}_{\text{o}}}\text{=}\frac{{\text{n}}_{\text{p}}}{{\text{n}}_{\text{s}}}\\ \therefore \frac{\text{2300 V}}{\text{230 V}}\text{=}\frac{\text{4000}}{{\text{n}}_{\text{s}}}\\ \therefore {\text{n}}_{\text{s}}\text{=}\frac{\text{4000 × 230}}{\text{2300}}\text{= 400}\\ \therefore \text{There are 400 turns in the secondary coil}\text{.}\end{array}

Q.23 Answer the following questions:

(a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?

(b) A capacitor is used in the primary circuit of an induction coil.

(c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.

(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.

(e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?

Ans.

(a) Yes; the statement is not true for rms voltage.

It is true that in any ac circuit, the applied instantaneous voltage is equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit. But, this is not true for rms voltage because such voltages across different elements may not be in same phase.

(b) The capacitor is used in the primary circuit of an induction coil. This is because, when a circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparking.

(c) The dc signal will appear across capacitor C because for dc signals, the impedance of an inductor (L) is very low while the impedance of a capacitor (C) is very high (almost infinite). The capacitor blocks the dc signal. Therefore, a dc signal appears across C. For an ac signal of high frequency, the impedance of L is high and that of C is negligible. The inductor blocks the ac signal of high frequency. Therefore, an ac signal of high frequency appears across L.

(d) On a dc line, choke offers no impedance. Therefore, the lamp shines brightly and the insertion of an iron core in the choke causes no change in the lamp’s brightness. However, on an ac line, both the choke coil and the iron core increase the impedance of the circuit. Therefore, if an iron core is inserted in the choke coil (which is in series with a lamp connected to the ac line), the lamp will glow dimly.

(e) A choke coil is needed in the use of fluorescent tubes with ac mains because it reduces ac across the tube without much loss of power. We cannot use an ordinary resistor instead of a choke coil for this purpose because it wastes power in the form of heat.

Q.24 At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m³s^–1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms^–2).

Ans.

$\begin{array}{l}\text{Here, height of water pressure head, h = 300 m}\hfill \\ {\text{Volume of the water flow per second, V = 100 m}}^{\text{3}}{\text{s}}^{\text{-1}}\hfill \\ \text{Efficiency of the turbine generator, η = 60\%}\hfill \\ \text{= 0}\text{.6}\hfill \\ \text{Acceleration due to gravity, g = 9}{\text{.8 ms}}^{\text{-2}}\hfill \\ {\text{Density of the water, ρ = 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\hfill \\ \text{Hydroelectric power =}\frac{\text{work}}{\text{time}}\hfill \\ \text{=}\frac{\text{Force × displacement}}{\text{time}}\hfill \\ \text{= Force × Velocity}\hfill \\ \text{=}\left(\text{Pressure × Area}\right)\text{× Velocity}\hfill \\ {\text{As, area×velocity = volumesec}}^{\text{-1}}\hfill \\ \text{= V}\hfill \\ \therefore \text{Hydroelectric power = P × V}\hfill \\ \text{=}\left(\text{hρg}\right)\text{× V}\hfill \\ \text{= hρgV}\hfill \end{array}$

Q.26 A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step up transformer at the plant.

Ans.

$\begin{array}{l}{\text{Here,\hspace{0.33em}total requirement\hspace{0.33em}of\hspace{0.33em}electric power, P = 800 kW = 800×10}}^{\text{3}}\text{W}\\ \text{Supply voltage, V = 220 V}\\ \text{Voltage at which the\hspace{0.33em}electric plant is generating power, V’ = 440 V}\\ \text{Distance between power generating station\hspace{0.33em}and\hspace{0.33em}the town, d =15 km}\\ \text{Resistance of the two wire lines carrying power = 0.5 Ω/km}\\ \text{Total resistance of the two\hspace{0.33em}wires, R = 2×15 km × 0.5Ω/km =15 Ω}\\ \text{As\hspace{0.33em}supply\hspace{0.33em}is\hspace{0.33em}through\hspace{0.33em}a step-down transformer of}\\ \text{rating 4000 – 220 V,}\\ \therefore {\text{Input voltage, V}}_{\text{i}}\text{= 4000 V}\\ {\text{Output voltage, V}}_{\text{o}}\text{= 220 V}\\ \text{Rms current in the wire lines,\hspace{0.33em}I =}\frac{\text{P}}{{\text{V}}_{\text{i}}}\text{=}\frac{{\text{800×10}}^{\text{3}}\mathrm{W}}{\text{4000 V}}\text{= 200\hspace{0.33em}A}\\ \left(\text{a}\right){\text{\hspace{0.17em}\hspace{0.17em}Line power loss = I}}^{\text{2}}\text{R}\\ \text{=}{\left(\text{200A}\right)}^{\text{2}}\text{×15}\mathrm{\Omega }{\text{= 600×10}}^{\text{3}}\text{W = 600 kW}\\ \left(\text{b}\right)\text{\hspace{0.17em}\hspace{0.17em}Assuming that there\hspace{0.33em}is\hspace{0.33em}negligible power loss\hspace{0.33em}due to leakage of current:}\\ \text{Essential\hspace{0.33em}power supplied by the electric\hspace{0.33em}power\hspace{0.33em}plant}\\ \text{= 800 kW+600 kW =1400 kW}\\ \left(\text{c}\right)\text{Voltage drop on the line = IR}\\ \text{=200 A ×15}\mathrm{\Omega }\text{=3000 V}\\ \therefore \text{Voltage transmitted from the\hspace{0.33em}power\hspace{0.33em}plant}\\ \text{= 3000 V + 4000 V =7000 V}\\ \text{As the power is\hspace{0.33em}generated at 440 V,}\\ \therefore \text{The rating of the step-up transformer situated at the power plant is 440 V-7000V.}\end{array}$

Q.27 Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.33em}}\mathrm{t}\text{he rating of a step}-\text{down transformer is}\\ \text{4}0000\text{\hspace{0.33em}V}-\text{22}0\text{V}.\\ \text{Input voltage},\\ {\text{V}}_{\mathrm{i}}=\text{4}0000\text{V}\\ \text{Output voltage},\\ {\text{V}}_{\mathrm{o}}=\text{22}0\text{V}\\ \text{Total requirement\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}electric power},\text{P}=\text{8}00\text{kW}\\ =\text{8}00\times \text{1}{0}^{\text{3}}\text{W}\\ \text{Source voltage},\text{V}=\text{22}0\text{V}\\ \text{Voltage at which the electric plant is\hspace{0.33em}generating power},\\ \text{V}‘=\text{44}0\text{V}\\ \text{Distance between power generating station\hspace{0.33em}and\hspace{0.33em}the}\\ \text{town},\text{d}=\text{15 km}\\ \text{The\hspace{0.33em}}\mathrm{r}\text{esistance of the two\hspace{0.33em}wire lines carrying power}\\ =0.\text{5}\mathrm{\Omega }{\text{km}}^{\text{-1}}\\ \text{Total resistance of wire lines},\\ \text{R}=\text{2}\times \text{15 km}\times 0.\text{5}\mathrm{\Omega }{\text{km}}^{\text{-1}}\\ \mathrm{R}=\text{15}\mathrm{\Omega }\\ \text{P}={\text{V}}_{\mathrm{i}}\text{I}\\ \\ \text{Rms current in the wire line,\hspace{0.33em}I=}\frac{\mathrm{P}}{{\text{V}}_{\mathrm{i}}}\\ =\frac{\text{8}00\times \text{1}{0}^{\text{3}}\mathrm{W}}{\text{4}0000\mathrm{V}}\\ =20\mathrm{A}\\ \left(\text{a}\right)\text{Line power loss}={\text{I}}^{\text{2}}\text{R}\\ ={\left(\text{2}0\right)}^{\text{2}}\times \text{15}\\ =\text{6 kW}\\ \left(\text{b}\right)\text{Assuming that there\hspace{0.33em}is negligible power loss\hspace{0.33em}due to}\\ \text{the leakage of current}.\\ \therefore \mathrm{P}\text{ower supplied by the plant}=\text{8}00\text{kW}+\text{6 kW}\\ =\text{8}0\text{6 kW}\\ \left(\text{c}\right)\text{Voltage drop on the line}=\text{IR}\\ =\text{2}0\mathrm{A}\times \text{15}\mathrm{\Omega }\\ =\text{3}00\text{V}\\ \therefore \text{Voltage transmitted by the power plant}\\ =\text{3}00\mathrm{V}+\text{4}0000\mathrm{V}\\ =\text{4}0\text{3}00\text{V}\\ \text{As\hspace{0.33em}}\mathrm{t}\text{he power is generated at 44}0\text{V,}\\ \therefore \text{The rating of the step}-\text{up transformer needed}\\ \text{at the plant is\hspace{0.33em}44}0\text{V}-\text{4}0\text{3}00\text{V}.\\ \therefore \text{Power loss during transmission}=\frac{6\mathrm{kW}}{806\text{kW}}\times 100\\ =0.744\mathrm{\%}\\ \text{In the previous exercise},\text{the power loss due to the}\\ \text{same reason =}\frac{600\mathrm{kW}}{1400\text{kW}}\times 100\\ =42.8\mathrm{\%}\\ \\ \therefore \text{High voltage transmission\hspace{0.33em}is preferred for this purpose.}\\ \end{array}$