NCERT Solutions For Class 10 Science Chapter 12 Electricity

(a)  Due to high melting point and high resistivity of tungsten, it does not burn at a high temperature. As the electric lamps glow at very high temperatures hence, the tungsten is used as heating element of electric bulbs.
(b)  Bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of pure metals. Due to high resistivity, they produce large amount of heat.
(c)  In a series circuit, there is a voltage distribution. Each element of a series circuit receives a small voltage for a large supply voltage. Thus, the amount of current decreases and the device becomes hot. Therefore, series arrangement is not used in domestic circuits.
(d)  Resistance (R) of a wire is inversely proportional to its area of cross-section (A).
(e)  The wires made of copper or aluminium have low resistivity and they are good conductors of electricity. Therefore, they are frequently used for electricity transmission.

$\begin{array}{l}\mathrm{Given},\text{ R = 8 }\mathrm{\Omega }\\ \text{ I = 15 A }\\ \mathrm{Rate}\text{ of heat produced can be expressed as power,}\\ {\text{ P = I}}^{\text{2}}\mathrm{R}\\ \mathrm{On}\text{ putting the values,}\\ \text{ P = }{\left(\text{15 A}\right)}^2\times 8=1800\text{ W or }1800\frac{\mathrm{J}}{\mathrm{s}}.\end{array}$

$\begin{array}{l}\text{Energy consumed by a TV set of power 250 W in 1 h}\\ \text{is given by the expression}\\ \text{ H = Pt = 250 W}\times \text{3600 s = 9}\times {\text{10}}^{\text{5}}\mathrm{J}\\ \mathrm{Similarly},\mathrm{energy}\text{ consumed by a toaster of power 1200 W}\\ \text{in 10 minutes = 1200}\times \text{600 = 7.2}\times {\text{10}}^{\text{5}}\mathrm{J}\\ \text{Hence, the energy consumed by a 250 W TV in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.}\end{array}$

$\begin{array}{l}\mathrm{The}\mathrm{potential}\text{ difference across each bulb will be 220}\\ \text{V because no division of voltage occurs in a parallel}\\ \text{circuit.}\\ \text{Current drawn by the bulb of rating 100 W is given by,}\\ {\text{ P = VI}}_{\text{1}}\\ \therefore {\text{ I}}_{\text{1}}\text{ = }\frac{\mathrm{P}}{\mathrm{V}}=\frac{100W}{220V}\\ \mathrm{Similarly},\text{ current drawn by the bulb of rating 60 W is}\\ \text{given by,}\\ {\text{ I}}_{\text{2}}\text{ = }\frac{\mathrm{P}}{\mathrm{V}}=\frac{60W}{220V}\\ \therefore \text{ Current drawn from the line = }\frac{100W}{220V}+\frac{60W}{220V}=0.727\mathrm{A}\end{array}$

$\begin{array}{l}\left(\mathrm{i}\right)\text{ Given, Potential difference, V = 6 V}\\ {\text{ R}}_{\text{1}}{\text{ = 1 Ω, R}}_{\text{2}}\text{ = 2 Ω}\\ \therefore {\text{ Equivalent resistance in series, R = R}}_{\text{1}}{\text{+ R}}_{\text{2}}\\ =\text{ (1+2) Ω = 3 Ω}\\ \text{ On using the relation,}\\ \text{ V = IR (I = current through the circuit)}\\ \\ \therefore \text{ I = }\frac{\text{V}}{\text{R}}\text{ = }\frac{\text{6}}{\text{3}}\text{ = 2 A}\\ \text{ Now, power in the circuit}\\ {\text{ P = I}}^{\text{2}}{\text{R = (2)}}^{\text{2}}\text{×2=8 W.}\\ \text{(ii) Given, potential difference, V = 4 V}\\ {\text{ R}}_{\text{1}}=\text{12 }\mathrm{\Omega },{\text{ R}}_2=2\mathrm{\Omega }\\ \text{ The voltage across each component of a }\\ \text{ parallel circuit remains the same. Hence, voltage across}\\ 2\mathrm{\Omega }\text{ resistor will be 4 V.}\\ \text{ Power consumed by 2 }\mathrm{\Omega }\text{ is given by}\\ \text{ P = }\frac{{\mathrm{V}}^2}{\mathrm{R}}=\frac{4^2}{2}=\text{ 8 W.}\end{array}$

$\begin{array}{l}\mathrm{Given},\text{ Supply voltage, V = 220 V}\\ \text{ Resistance, R = 24 }\mathrm{\Omega }\\ \text{(i) When coils are used separately:}\\ \text{ By using the relation,}\\ {\text{ V = I}}_{\text{1}}{\text{R}}_{\text{1}}\\ {\text{ Where, I}}_{\text{1}}\text{ = current flowing through the coil}\\ \therefore {\text{ I}}_{\text{1}}\text{= }\frac{\text{V}}{{\text{R}}_{\text{1}}}\text{ = }\frac{\text{220}}{\text{24}}\text{ = 9.166 A.}\\ \\ \text{(ii) When coils are connceted in series:}\\ \text{ total resistance of the two coils}\\ {\text{ R}}_{\text{2}}\text{ = (24+24) }\mathrm{\Omega }\text{ = 48 }\mathrm{\Omega }\\ \text{ By using the relation,}\\ {\text{ V = I}}_2{\text{R}}_2\\ \therefore {\text{ I}}_{\text{2}}\text{= }\frac{\text{V}}{{\text{R}}_{\text{2}}}\text{ = }\frac{\text{220}}{\text{48}}\text{ = 4.58 A.}\\ \text{(iii) When coils are connceted in parallel:}\\ {\text{ Total resistance (R}}_{\text{3}})\text{ would be}\\ \frac{1}{{\text{R}}_{\text{3}}}=\frac{1}{24}+\frac{1}{24}=\frac{2}{24}\\ \mathrm{or},{\text{ R}}_{\text{3}}=12\mathrm{\Omega }\\ \text{ On using the relation}\\ {\text{ V = I}}_3{\text{R}}_3\\ \therefore {\text{ I}}_3=\frac{\mathrm{V}}{{\text{R}}_3}=\frac{220}{12}=18.33\text{ A.}\end{array}$

$\begin{array}{l}\mathrm{Given},\text{ V = 220 V}\\ \text{ i = 5 A}\\ {\text{ P}}_{\text{1}}\text{ = 10 W}\\ \text{On using the relation,}\\ \text{ P = }\frac{{\mathrm{V}}^2}{{\mathrm{R}}_1}\\ {\text{or, R}}_{\text{1}}\text{ = }\frac{{\mathrm{V}}^2}{{\mathrm{P}}_1}\\ \therefore \text{ = }\frac{{\left(220\right)}^2}{10}\text{= 4840 }\mathrm{\Omega }\\ \text{As per ohm's law,}\\ \text{ V= IR }\\ \therefore \text{ R =}\frac{\mathrm{V}}{\mathrm{I}}\text{= }\frac{220}{5}=44\mathrm{\Omega }\\ \mathrm{Resistance}\text{ of each electric bulb,}\\ {\text{ R}}_{\text{1}}=4840\mathrm{\Omega }\\ \therefore \frac{1}{\mathrm{R}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_1}+....+\mathrm{upto}\text{ x times}\\ \text{or, x = }\frac{{\mathrm{R}}_1}{\mathrm{R}}=\frac{4840}{44}=110\\ \mathrm{Thus},\text{ 110 electric bulbs are connected in parallel.}\end{array}$

(i) 


$\begin{array}{l}\text{When two resistors are connected in parallel and}\\ \text{the third one is connected in series with these}\\ \text{two then, Two 6 }\mathrm{\Omega }\text{ resistors are connected in parallel.}\\ \text{Their equivalent resistance (R') will be}\\ \frac{1}{\mathrm{R}\text{'}}=\frac{1}{6\mathrm{\Omega }}+\frac{1}{6\mathrm{\Omega }}=\frac{2}{6\mathrm{\Omega }}=\frac{1}{3\mathrm{\Omega }}\\ \\ \text{ or, R' = 3 }\mathrm{\Omega }\\ \text{The third 6 }\mathrm{\Omega }\text{ resistor is connected in series with 3 }\mathrm{\Omega }.\\ {\text{Hence, the equivalent resistance R}}_{\text{1}}\text{ = (6+3) }\mathrm{\Omega }\text{ = 9 }\mathrm{\Omega }\end{array}$

(ii)


$\begin{array}{l}\text{Two resistors of 6 Ω are in series. Their equivalent}\\ \text{resistance will be = (6}+6\text{) Ω = 12 Ω}\\ \text{The third 6 Ω resistor is connected in parallel with 12 Ω.}\\ {\text{Hence, equivalent resistance (R}}_{\text{2}}\text{)}\\ \frac{\text{1}}{{\text{R}}_{\text{2}}}\text{=}\frac{\text{1}}{\text{12\hspace{0.17em}Ω}}\text{+}\frac{\text{1}}{\text{6\hspace{0.33em}Ω}}\text{=}\frac{\text{3}}{\text{12\hspace{0.33em}Ω}}\text{=}\frac{\text{1}}{\text{4\hspace{0.33em}Ω}}\\ {\text{or, R}}_{\text{2}}\text{ = 4 Ω}\end{array}$

$\begin{array}{l}\text{The current would be same throughout the series circuit.}\\ \text{Equivalent resistance of the circuit,}\\ {\text{R}}_{\text{eq}}=(0.2+0.3+0.4+0.5+12)\mathrm{\Omega }=13.4\mathrm{\Omega }\\ \mathrm{V}=\text{ 9 V}\\ \text{By using ohm's law,}\\ \text{ V =IR}\\ \mathrm{or},\text{ I =}\frac{\mathrm{V}}{\mathrm{R}}\\ \therefore \text{ I = }\frac{9}{13.4\mathrm{\Omega }}\text{ = 0.671 A}\\ \text{Therefore, the current that would flow through}\\ \text{12 }\mathrm{\Omega }\text{ resistor is 0.671 A. }\end{array}$

$\begin{array}{l}\mathrm{According}\text{ to ohm's law, }\\ \text{ V = IR}\\ \text{ or, R = }\frac{\mathrm{V}}{\mathrm{I}}\\ \text{Here, potential difference, V = 12 volts}\\ \text{ i = 2.5 mA}\\ \text{ =2.5 }\times {\text{10}}^{\text{-3}}\text{ A}\\ \therefore \text{ R = }\frac{12}{2.5\times {\text{10}}^{\text{-3}}}\text{=4.8 }\times {\text{10}}^{\text{-3}}\mathrm{\Omega }.\end{array}$

When the graph is plotted between voltage and current then, it is called V-I characteristics. Here, the voltage is plotted on X axis and current is plotted on Y axis.



$\begin{array}{l}\mathrm{The}\text{ slop of the line gives the value of resistance (R)}\\ \text{as,}\\ \text{Slope, =}\frac{1}{\mathrm{R}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{2}{6.8}\\ \therefore \mathrm{R}=\frac{6.8}{2}=3.4\mathrm{\Omega }.\end{array}$

A voltmeter need to be connected in parallel in a circuit to measure the potential difference between two points.

$\begin{array}{l}\text{The correct option is }\left(\text{c}\right).\\ \text{Suppose, R = resistance of each wire}\\ \therefore \text{ The equivalent resistance in series }\\ {\text{ R}}_{\text{s}}\text{ = R+R = 2R}\\ \text{If V is the applied potential difference, then it is the}\\ \text{voltage across the equivalent resistance.}\\ {\text{V = I}}_{\text{s}}\times {\text{2R (I}}_{\text{s}}\text{= current in series combination)}\\ \\ \Rightarrow {\text{ I}}_{\text{s}}\text{ = }\frac{\mathrm{V}}{2\mathrm{R}}\\ \mathrm{The}\text{ heat dissipated in time t is,}\\ {\text{H}}_{\text{s}}={\text{ I}}_{\mathrm{s}}^2\times 2\mathrm{R}\times \mathrm{t}={\left(\frac{\mathrm{V}}{2\mathrm{R}}\right)}^2\times 2\mathrm{R}\times \mathrm{t}\\ \Rightarrow {\text{ H}}_{\text{s}}=\frac{{\mathrm{V}}^2\mathrm{t}}{2\mathrm{R}}...\text{ (i)}\\ \mathrm{The}\text{ equivalent resistance of the parallel connection is}\\ {\text{R}}_{\text{p}}=\frac{1}{\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}}=\frac{\mathrm{R}}{2}\\ \mathrm{V}{\text{ is the applied potential difference across R}}_{\text{p}}.\\ \therefore \mathrm{V}={\text{I}}_{\mathrm{p}}\times \frac{\mathrm{R}}{2}\end{array}$

$\begin{array}{l}{\text{Given, V}}_{\text{1}}{\text{ = 220 V, V}}_{\text{2}}=110\text{ V}\\ {\text{ P}}_{\text{1}}=\text{ 100 W}\\ \mathrm{On}\text{ using the relation,}\\ {\text{ P}}_{\text{1}}\text{ = }\frac{{\text{V}}_1^2}{\mathrm{R}}\\ \text{or, R = }\frac{{\mathrm{V}}_1^2}{{\mathrm{P}}_1}\text{ = }\frac{{\left(220\text{ V}\right)}^2}{\left(100\text{ W}\right)}=484\mathrm{\Omega }\\ \text{If the bulb is operated on 110 V, then the energy}\\ \text{consumed,}\\ {\text{ P}}_{\text{2}}=\frac{{\text{V}}_2^2}{\mathrm{R}}=\frac{{\left(110\text{ V}\right)}^2}{484\mathrm{\Omega }}=\text{ 25 W}\\ \text{Therefore, the power consumed will be 25 W.}\end{array}$

\begin{array}{l}\text{Electrical power is given by,}\\ \text{ P= VI }\mathrm{...}\text{ (1)}\\ \text{According to ohm's law,}\\ \text{ V = IR }\mathrm{...}\text{ (2)}\\ \text{Where, V = potential difference}\\ \text{ I = current }\\ \text{ R = resistance}\\ \therefore \text{ P = VI}\\ \text{From equation (1), P = (IR)×I}\\ \therefore {\text{ P = I}}^{\text{2}}\text{R}\\ \text{From equation (2),}\\ \text{ I = }\frac{\text{V}}{\text{R}}\\ \therefore \text{ P =}\frac{{\text{V}}^{\text{2}}}{\text{R}}\\ {\text{Hence, P = VI = I}}^{\text{2}}\text{R =}\frac{{\text{V}}^{\text{2}}}{\text{R}}\\ {\text{Thus, IR}}^{\text{2}}\text{ does not represent electrical power}\text{.}\end{array}

The correct option is (d).

$\begin{array}{l}\text{The resistance of a piece of wire is }\mathrm{R}\text{ and it is proportional}\\ \text{to the length of wire. As the wire is cut into five equal parts }\\ \text{h}\mathrm{ence},\text{ resistance of each part = }\frac{\mathrm{R}}{5}\\ \text{Let the equivalent resistance of these piece of wire in}\\ \text{parallel is R' then,}\\ \frac{1}{\mathrm{R}\text{'}}=\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}=\frac{25}{\mathrm{R}}\\ \mathrm{or},\frac{\mathrm{R}}{\mathrm{R}\text{'}}\text{ = 25}\end{array}$