# Important Questions Class 12 Physics Chapter 7

## Important Questions for CBSE Class 12 Physics Chapter 7 – Alternating Current

These important questions for class 12 physics chapter 7 will teach students about alternating current. After reviewing these CBSE revision notes, students will understand what alternating current is. Students would be better prepared for their exams by studying these chapter 7 class 12 physics important questions and would be able to solve several CBSE sample papers after studying these physics class 12 chapter 7 important questions.

These notes also come with important formulas and CBSE additional questions that can be used to assess your comprehension.

### Study Important Questions for Class 12 Physics Chapter 7 – Alternating Current

Here are some important questions that students can study to prepare better for their examinations.

### Important Questions For Class 12 Physics Chapter 7

1. If an a.c. the circuit has a power factor of 0.5, what will the phase difference between the circuit’s voltage and current be?

Ans. Cos -0.5- 1/2

= 60

Hence, the phase difference is 60.

1. Define the physical quantity represented by the Weber unit.

Ans. The SI unit for magnetic flux is the Weber.

One Weber is the magnetic flux that links a circuit of one turn, which would produce in it an emf of 1 volt if it were reduced to 0 at a uniform rate in 1 second.

1. In the same magnetic field, two identical loops—one made of copper and the other of aluminium—rotate at the same pace. Which scenario will result in a higher induced

(a) emf?

(b) current, and why?

Ans. The induced electromotive force will be the same in both loops, but the copper loop’s induced current will be higher due to its lower resistance.

1. Can D.C. voltage not be increased using a transformer?

Ans. The current in the transformer’s primary coil doesn’t change when a d.c. the voltage source is applied across it. As a result, the magnetic flux connected to the secondary remains unchanged. And there is no voltage across the secondary coil. Hence, a transformer cannot increase dc voltage.

1. As illustrated in the figure, the magnet is moved between coils AB and CD in the direction indicated by the arrow. Indicate which way the electricity in each coil is flowing.

Ans.  (1) Impedance total across the circuit

Z= R 2+ (XL – XC)2

(2) f = 12 LC

1. When does a coil’s self-induction change?

(1) A coil’s number of turns is reduced.

(2) A rod made of iron is inserted into it. Justify.

Ans. L = 2H, C = 32 , F= 32 X 10 -6F, R= 10

WR = 1LC = 12 X 32 X 10 -6

= – Value of the circuit

= 1R LC = 110  232 X 10-5

= 25

1. An alternating source of emf E = is connected in series with a coil of inductance L, capacitance C, and resistance R. (Eo sin wt). Write an expression for the:

(1) circuit’s overall impedance.

(2) source emf frequency at which the current-carrying circuit exhibits resonance.

Ans. (1) Total impedance of the circuit

Z = R  2 + (XL – Xc) 2

(2) f= 12LC

1. Find the resonant frequency, w, of a series LCR circuit with L = 2 H, C = 32 F, and R = 10 . What is the – value of this circuit?

Ans. L = 2H

C = 32

F = 32 X 10-6F

R = 10

– Value of the circuit

= 1R LC = 110 232 X 10-5

= 25

1. A 220 V, 50 Hz ac supply is coupled with a 100 resistor.

(a) What is the RMS value of the circuit’s current?

(b) What is the total net power used during a cycle?

Ans. R = 100, which is the resistor’s resistance. The supply voltage is 220 volts, and the frequency is 50 Hz.

(a) The circuit’s RMS value of current is given as

l =VR

= 220100

= 2.20 A

(b) The following is the net power used during a full cycle:

P = VI

= 220 X 2.2

= 484 W

1. (a) Ac supplies have a peak voltage of 300 V. What voltage is the RMS?

(b) An ac circuit’s RMS current value is 10 A. Which current is at its peak?

Ans. (a) Peak voltage of the AC supply, V= 300V

RMS voltage is given as

V = V2 = 3002 = 212.1V

(b) RMS value of current is:

I = 10A

l= 2l

= 102

= 14.1 A

1. A 220 V, 50 Hz ac supply is coupled to a 44 mH inductor. Find the current in the circuit’s RMS value.

Ans. The inductance of inductor

L = 44 mH = 44 X 10-3 H

Supply voltage, V = 220V

Frequency, = 50 Hz

Angular frequency, = 2v

Inductive reactance, XL = XL=L=2L = 2 X 50 X 44 X 10-3

RMS value of current is given as l = vXL s

= 2002 X 50 X 44 X 10-3  = 15.92 A.

1. A 110 V, 60 Hz ac supply is coupled with a 60F capacitor. Find the current in the circuit’s RMS value.

Ans. The capacitance of the capacitor, C = 60 F = 60 X 10-6 F

Supply voltage, V= 110 V

Frequency, = 60Hz

Angular frequency, =2

Capacitive reactance XC = 1(i) C

= 12C

= 12 X 3.14 X 60 X 60X 10-6 -1

RMS value of current is given as

L = Xc

= 110 X 2 X 3.14 X 60 X 10-6 X 60

=  2.49 A.

1. Determine the series LCR circuit’s resonance frequency with an R = 10, L= 2.0 H and C= 32 F. What is this circuit’s Q-value?

Ans. L = 2.0 H

C = 32 F  =  32 X 10-6F

R = 10

Resonating frequency by the relation:

(i) r = 1LC

= 12 X 32 X 10-6

= 18 X 10-3

= 125 S -1

The circuit’s Q-value is now reported as

Q= 1R LC

= 110 232 X 10-6

= 110 X 14 X 103

= 25

1. The variation of an alternating emf with time is depicted in the image given below. What is the average emf value for the area of the graph that is shaded?

Ans. Em = 0.637, Eo = 0.637 314 or Em = 200 V is the average or mean value of ac throughout a half-cycle or in time T/2.

1. What power is lost in an ac circuit when the voltage and current are

V = 230 sin(t + 2/2) and l = 10 sin t, respectively?

Ans. Power consumed = Vrms lrms cos /2 = 0 since the voltage and current have a phase difference of /2.

1. When a lamp is linked to an AC power source, it illuminates with an equivalent amount of brightness to a 12 V DC battery. What is the alternating voltage’s peak value?

Ans. The peak value is V = 12 × 2

= 16.97 V

1. A light bulb is connected in series with an ac source (a) across an LR circuit and (b) across an RC circuit. Give a mathematical formula to illustrate how increasing the ac source frequency affects the bulb’s brightness in cases (a) and (b).

Ans. (a) The source of the current in the LR circuit is

I = VR2+ 2 L2

Brightness diminishes as l lowers and the frequency of the ac source ω increases.

(b) The current in the RC circuit is given by

I = VR2+ 12 C2

Increases in l and brightness result from an increase in the frequency of the ac source.

1. An ac source, a light, and an air-core solenoid are all linked. What happens to the brightness of the bulb if an iron core is put inside the solenoid? Explicitly justify your response.

Ans. The solenoid’s inductance is increased by inserting an iron core. Inductive reactance then rises in value as a result of this. Therefore, the bulb’s brightness drops along with the current.

1. Mention the variables that affect a series LCR circuit’s resonant frequency. Plot a graph demonstrating how the frequency of the applied ac source affects the impedance of a series LCR circuit.

Ans. The amount of inductance L and capacitance C in a series LCR circuit determines its resonance frequency.

The graph below shows how the applied ac source’s frequency, f, affects the impedance Z of a series LCR circuit.

1. A capacitor and a light are linked in series. For dc and ac connections, forecast your observations. What happens if the capacitor’s capacitance is decreased in each scenario?

Ans. When a capacitor is linked to a DC source, the capacitor charges. Once charged, no current flows through the circuit, and the lamp won’t light up. Even if C is decreased, nothing will have changed. The capacitor provides capacitive reactance (1/C) in the presence of an ac source, and the circuit’s current flows. As a result, the lamp will be illuminated. Reducing C will result in an increase in reactance and a decrease in light brightness.

### Alternating Current Important Questions

1. The secondary transformer, whose primary draw line voltage is wired to a 60 W load. What is the primary coil’s current if the load has a 0.54 A current flow? Describe the kind of transformer that is being used.

Ans. Ps = 60 W, and ls = 0.54 A

Given that P=V I, we have Vs = Ps/ls, which is 60/0.54 = 110 V.

The transformer is, therefore, a step-down transformer, and its transformation ratio is 1/2. As a result, lp = 1/2 ls = 1/2 0.54 0.27

1. A little community 15 kilometres from a 440 V electric producing station with a requirement for 800 kW of 220 V electricity. The two wirelines carrying power have a resistance of 0.5 per km. A substation in the town houses a 4000 – 220 V step-down transformer that provides the town with power from the line.

(a) Calculate the heat-related line power loss.

(b) Given a minimal power loss from leakage, how much electricity must the plant produce?

(c) Describe the plant’s step-up transformer.

Ans. Electricity is necessary P = 800 kW = 800 × 103 W

Required voltage V0 = 220 V

The power station is 15 kilometres away or 15 103 metres.

Kilometres resistance equals 0.5 ohms per kilometre

Total resistance, R, is equal to 0.5 2 15 ohms.

The voltage at input Vi = 440 V

Transformer’s primary voltage εp is 4000 V.

Transformer’s secondary voltage εs is 220 V.

(a) The two-wire line’s RMS current value

lrms = Prms = 800 X 1034000 = 200 A

hence, line-wide power loss = l2rmsR = (200)2 × 15 = 600 kW

(b) Assuming there will be little loss from leakage, total power supply equals town power demand plus line power loss

P = 800 kW + 600 kW = 1400 kW

(c)  Voltage drop on the line is equals 200 15 = 3000 V.

Consequently, the generation station’s voltage is V=4000 + 3000 = 7000 V.

As a result, the plant’s step-up transformer has a voltage range of 440 V to 7000 V.

1. An ac generator is made up of two fixed pole pieces and a coil with 50 turns and a surface area of 2.5 m2 that rotates in a uniform magnetic field B = 0.2 tesla at an angle of 60 rads-1. The circuit’s resistance, including the resistance of the coil, is 500 ohms.

(a) Determine the generator’s maximum current draw.

(b) What is the flux at zero current?

(c) Would the generator still function if the coil were stationary, but the pole pieces spun at the same rate as before? Justify your response.

Ans. Given n = 50, A = 2.5 m2, ω = 60 rads-1, B = 0.2 T, R = 500 ohm, l0 = ?

(a) The coil’s maximum amount of induced emf is

ε0 = nBAω = 50 × 0.2 × 2.5 × 60 = 1500 V

Consequently, the circuit’s maximum current is

l0 = 0R = 1500500 = 3A

(b)The maximum flux in this instance is provided by Φ = nBA = 50 × 0.2 × 2.5 = 25 Wb.

(c) Yes, the magnet and coil must move relative to one another for a generator to function.

1. A 220 V source is coupled with a pure inductor measuring 25.0 mH. If the source’s frequency is 50 Hz, calculate the circuit’s inductive reactance and peak current.

Ans. The inductive reactance,

XL = 2 πf L = 2 × 3.14 × 50 × 25 × 10-3 Ω

= 7.85 ohm

The circuit’s RMS current is

I =  VXL= 2207.85 = 28 A

1. Describe the operation and principle of an ac generator using a diagram.

Ans. It is founded on the electromagnetic induction theory. An induced emf is generated across a coil when it is turned approximately parallel to the direction of a homogeneous magnetic field.

Working: As indicated in the image below, the various armature positions can be used to understand how the ac generator operates.

Assume that at time t = 0, the loop’s plane is parallel to B. The induced emf changes from zero to its maximum value and then back to zero when the loop turns around in place t = 0 to point t T/2, as depicted in the diagram. Magnetic flux has no instantaneous rate and is zero for angles 0° and 1800; hence, there is no induced emf. The emf again oscillates between zero and its highest value before returning to zero when the Loop advances between the positions t = T/2 and t = T. But this time, it changes course. The maximum magnetic flux is coupled with the coil at angles of 90° and 270°. Hence the emf is greatest at these angles. As a result, the AC generator’s output varies sinusoidally with time.

1. Give the basic operating principles of an AC generator and use a labelled diagram to demonstrate. Find the expression for the induced electromagnetic field (EMF) in a coil with N turns and cross-sectional area A that is rotating perpendicular to the axis of rotation in a magnetic field B.

Ans. For a theory and a diagram,

Assume that at time t = 0, the loop’s plane is parallel to B. The Induced emf changes from zero to its maximum value and then back to zero when the loop turns around in place t = 0 to point t T/2, as depicted in the diagram. Magnetic flux has no instantaneous rate and is zero for angles 0° and 1800; hence, there is no induced emf. Returning to zero, the emf increases to its maximum value, then decreases to zero when the Loop advances between the points t = T/2 and t = T.  But this time, it changes course. The maximum magnetic flux is coupled with the coil at angles of 90° and 270°. Hence the emf is greatest at these angles. As a result, the AC generator’s output varies sinusoidally with time.

The emf it produced is determined by the equation ε = nBAω Sin ωt, where ω is the coil’s rotational speed.

1. A plane travels horizontally from west to east at 900 kilometres per hour. Determine the probable difference that could emerge between the wings’ 20-meter-long ends. The Earth’s magnetic field has a horizontal component of 5 × 10-4 T, and the dip angle is 30°.

Ans. The potential difference between the wingtips emerged

ε = Blv

Given,

Velocity = 900 km h-1 = 250 m s-1

Wingspan of (L1) = 20 m

Magnetic field of the Earth’s vertical component

Bv = BH tan δ = 5 × 10-4 (tan 30°) T

Hence, pd is

= 5 × 10-4 (tan 30°) × 20 × 250 = 1.44 V

1. Create a diagram of a device that converts high ac voltage to low ac voltage and describe its operation. In this gadget, list four potential sources of energy loss.

Ans. As shown in the labelled diagram.

According to the mutual induction theory, whenever the magnetic flux associated with one coil changes, a surrounding coil experiences an induced emf.

The following are potential reasons for energy losses in transformers:

• Flux leakage
• Copper loss
• Eddy currents
• Hysteresis loss
1. A tiny hamlet 20 kilometres from a 440V electric producing station has a demand for 1200kW of electric electricity at 220 V. The two wirelines carrying power have a resistance of 0.5 per km. At a substation in the town, a 4000 – 220V step-down transformer provides the town with power from the line. Calculate the heat-related line power loss.

Ans. Total resistance = length × resistance per unit length = 40 km × 0.5 = 20 Ω

Flowing current in the line l = P/V

= 1200 × 103 /4000 = 300 A

Power loss would be = l2R = (300)2 × 20 = 1800 kW

1. A capacitor and a light are linked in series. When connected in turn across, this combination of (i) an ac source and (ii) a “dc” battery, forecast your observation. What difference would you observe if the capacitor’s capacitance was raised in each scenario?

Ans. The condenser charges when a dc source is attached, but no current travels through the circuit; as a result, the lamp does not light up. Even when the capacitor’s capacitance is increased, nothing changes.

The capacitor provides capacitive reactance when an ac source is attached XC=1C. The circuit is conducting current, and the lamp is lit. Xc drops as capacitance rises. Therefore, the glow of the bulb increases.

### Conclusion:

To help students with exam preparation, Extramarks provides Class 12 Physics Chapter 7 important questions. Extramarks also provides the best quality CBSE previous year question paper based on NCERT books and according to the latest CBSE syllabus. The questions are well-structured and cover all the important concepts. These can be of great help in last-minute revision.

Q1- A generator produces a voltage V=240sin120t volt. The value of frequency and rms voltage is

Opt-

A-19 Hz and 120 V

b-19 Hz and 170 V

c-60 Hz and 240 V

d-754 Hz and 170 V

Ans- ## 1. Give a definition of "wattless current".

Wattless current is the portion of circuit current that results in zero power consumption in the circuit.

## 2. Explain the quality factor in an LCR series circuit. What is its SI equivalent?

The ratio of the voltage created across the capacitor or inductor to the applied voltage is known as the quality factor and lacks any kind of unit.

## 3. Can a capacitor with the right capacitance replace a choke?

Yes, because even then, there won’t be any power used.

One.