NCERT Solutions for Class 12 Physics Chapter 7: Alternating Current

An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where,

L = 5.0 H

C = 80 μF = 80 × 10−6 F

R = 40 Ω

Potential of the voltage source, V = 230 V

Impedance (Z) of the given parallel LCR circuit is given as:

Where,

ω = Angular frequency

At resonance,

Hence, the magnitude of Z is the maximum at 50 rad/s. As a result, the total current is minimum.

Rms current flowing through inductor L is given as:

Rms current flowing through capacitor C is given as:

Rms current flowing through resistor R is given as:

 

Inductance, L = 80 mH = 80 × 10−3 H

Capacitance, C = 60 μF = 60 × 10−6 F

Supply voltage, V = 230 V

Frequency, ν = 50 Hz

Angular frequency, ω = 2πν= 100 π rad/s

Peak voltage, V0 =

(a) Maximum current is given as:

The negative sign appears because

Amplitude of maximum current,

Hence, rms value of current,

(b) Potential difference across the inductor,

VL= I × ωL

= 8.22 × 100 π × 80 × 10−3

= 206.61 V

Potential difference across the capacitor,

(c) Average power consumed by the inductor is zero as actual voltage leads the current by.

(d) Average power consumed by the capacitor is zero as voltage lags current by.

(e) The total power absorbed (averaged over one cycle) is zero.

 

Resistance of the resistor, R = 100 Ω

Supply voltage, V = 220 V

Frequency, ν = 50 Hz

(a) The rms value of current in the circuit is given as:

(b) The net power consumed over a full cycle is given as:

P = VI

= 220 × 2.2 = 484 W

 

(a) Peak voltage of the ac supply, V0 = 300 V

Rms voltage is given as:

(b) Therms value of current is given as:

I = 10 A

Now, peak current is given as:

 

As given : Inductance of inductor, L = 44 m H = 44 × 10 – 3 H

Voltage of source , V = 220 V

Frequency of source , ν = 50 Hz

Angular frequency of source , ω = 2 π ν

à  Inductive reactance, X L = ω L = 2 π ν L = 2π × 50 × 44 × 10 – 3 Ω

We know that :

Rms value of current : à  I

Hence, the rms value of current in the circuit is 15.92 A.

 

Capacitance of capacitor, C = 60 μF = 60 × 10−6 F

Supply voltage, V = 110 V

Frequency, ν = 60 Hz

Angular frequency, ω=

Capacitive reactance

Rms value of current is given as:

Hence, the rms value of current is 2.49 A.

 

Given that :  Inductive network

In the above circuit we have :

rms current value, I = 15.92 A

rms voltage value, V = 220 V

Therefore, the total power taken in can be derived by the following equation :

à  P = VI cos Φ

Here ,

Φ = Phase difference between V and I.

We know that, the difference in phase of alternating voltage and alternating current is 90° , in case of a pure inductive circuit

i.e., Φ = 90°.

Therefore, P = 0

i.e., the total power is zero.

In case of the capacitive network,

The value of rms current is given by , I = 2.49 A

The value of rms voltage is given by , V = 110 V

Thus , the total power taken in can be derived from the following equation :

à  P = VI Cos Φ

For a pure capacitive circuit, the phase difference between alternating Voltage and alternating current is 90°

i.e., Φ = 90°.

Thus , P = 0

i.e., the net power is zero.

 

Height of water pressure head, h = 300 m

Volume of water flow per second, V = 100 m3/s

Efficiency of turbine generator, n = 60% = 0.6

Acceleration due to gravity, g = 9.8 m/s2

Density of water, ρ = 103 kg/m3

Electric power available from the plant = η × hρgV

= 0.6 × 300 × 103 × 9.8 × 100

= 176.4 × 106 W

= 176.4 MW

 

Inductance, L = 2.0 H

Capacitance, C = 32 μF = 32 × 10−6 F

Resistance, R = 10 Ω

Resonant frequency is given by the relation,

Now, Q-value of the circuit is given as:

Hence, the Q-Value of this circuit is 25.

 

Capacitance, C = 30μF = 30×10−6F

Inductance, L = 27 mH = 27 × 10−3 H

Angular frequency is given as:

Hence, the angular frequency of free oscillations of the circuit is 1.11 × 103 rad/s.

 

Capacitance of the capacitor, C = 30 μF = 30×10−6 F

Inductance of the inductor, L = 27 mH = 27 × 10−3 H

Charge on the capacitor, Q = 6 mC = 6 × 10−3 C

Total energy stored in the capacitor can be calculated by the relation,

Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.

 

At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit.

Resistance, R = 20 Ω

Inductance, L = 1.5 H

Capacitance, C = 35 μF = 30 × 10−6 F

AC supply voltage to the LCR circuit, V = 200 V

Impedance of the circuit is given by the relation,

At resonance,

Current in the circuit can be calculated as:

Hence, the average power transferred to the circuit in one complete cycle= VI

= 200 × 10 = 2000 W.

 

The range of frequency (ν) of a radio is 800 kHz to 1200 kHz.

Lower tuning frequency, ν1 = 800 kHz = 800 × 103 Hz

Upper tuning frequency, ν2 = 1200 kHz = 1200× 103 Hz

Effective inductance of circuit L = 200 μH = 200 × 10−6 H

Capacitance of variable capacitor for ν1 is given as:

C1

Where,

ω1 = Angular frequency for capacitor C1

Capacitance of variable capacitor for ν2,

C2

Where,

ω2 = Angular frequency for capacitor C2

Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.

 

Inductance of the inductor, L = 5.0 H

Capacitance of the capacitor, C = 80 μF = 80 × 10−6 F

Resistance of the resistor, R = 40 Ω

Potential of the variable voltage source, V = 230 V

(a) Resonance angular frequency is given as:

Hence, the circuit will come in resonance for a source frequency of 50 rad/s.

(b) Impedance of the circuit is given by the relation,

At resonance,

Amplitude of the current at the resonating frequency is given as:

Where,

V0 = Peak voltage

Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.

(c) Rms potential drop across the inductor,

(VL)rms = I × ωRL

Where,

I = rms current

Potential drop across the capacitor,

Potential drop across the resistor,

(VR)rms = IR

=

Potential drop across the LC combination,

At resonance,

∴VLC= 0

Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.

 

Inductance of the inductor, L = 20 mH = 20 × 10−3 H

Capacitance of the capacitor, C = 50 μF = 50 × 10−6 F

Initial charge on the capacitor, Q = 10 mC = 10 × 10−3 C

(a) Total energy stored initially in the circuit is given as:

Hence, the total energy stored in the LC circuit will be conserved because there is no resistor connected in the circuit.

(b)Natural frequency of the circuit is given by the relation,

Natural angular frequency,

Hence, the natural frequency of the circuit is 103 rad/s.

(c) (i) For time period (T), total charge on the capacitor at time t,

For energy stored is electrical, we can write Q’ = Q.

Hence, it can be inferred that the energy stored in the capacitor is completely electrical at time, t =

(ii) Magnetic energy is the maximum when electrical energy, Q′ is equal to 0.

Hence, it can be inferred that the energy stored in the capacitor is completely magnetic at time,

(d) Q1 = Charge on the capacitor when total energy is equally shared between the capacitor and the inductor at time t.

When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor = (maximum energy).

Hence, total energy is equally shared between the inductor and the capacity at time,

(e) If a resistor is inserted in the circuit, then total initial energy is dissipated as heat energy in the circuit. The resistance damps out the LC oscillation.

 

Inductance of the inductor, L = 0.50 H

Resistance of the resistor, R = 100 Ω

Potential of the supply voltage, V = 240 V

Frequency of the supply, ν = 50 Hz

(a) Peak voltage is given as:

Angular frequency of the supply,

ω = 2 πν

= 2π × 50 = 100 π rad/s

Maximum current in the circuit is given as:

(b) Equation for voltage is given as:

V = V0 cos ωt

Equation for current is given as:

I = I0 cos (ωt − Φ)

Where,

Φ = Phase difference between voltage and current

At time, t = 0.

V = V0(voltage is maximum)

Forωt − Φ = 0 i.e., at time,

I = I0 (current is maximum)

Hence, the time lag between maximum voltage and maximum current is.

Now, phase angle Φis given by the relation,

Hence, the time lag between maximum voltage and maximum current is 3.2 ms.

 

Inductance of the inductor, L = 0.5 Hz

Resistance of the resistor, R = 100 Ω

Potential of the supply voltages, V = 240 V

Frequency of the supply,ν = 10 kHz = 104 Hz

Angular frequency, ω = 2πν= 2π × 104 rad/s

(a) Peak voltage,

Maximum current,

(b) For phase differenceΦ, we have the relation:

It can be observed that I0 is very small in this case. Hence, at high frequencies, the inductor amounts to an open circuit.

In a dc circuit, after a steady state is achieved, ω = 0. Hence, inductor L behaves like a pure conducting object.

 

Capacitance of the capacitor, C = 100 μF = 100 × 10−6 F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

(a) Frequency of oscillations, ν= 60 Hz

Angular frequency,

For a RC circuit, we have the relation for impedance as:

Peak voltage,

Maximum current is given as:

 

(b) In a capacitor circuit, the voltage lags behind the current by a phase angle ofΦ. This angle is given by the relation:

Hence, the time lag between maximum current and maximum voltage is 1.55 ms.

 

Capacitance of the capacitor, C = 100 μF = 100 × 10−6 F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

Frequency of the supply, ν = 12 kHz = 12 × 103 Hz

Angular Frequency, ω = 2 πν= 2 × π × 12 × 10303

= 24π × 103 rad/s

Peak voltage,

Maximum current,

For an RC circuit, the voltage lags behind the current by a phase angle of Φ given as:

Hence, Φ tends to become zero at high frequencies. At a high frequency, capacitor C acts as a conductor.

In a dc circuit, after the steady state is achieved, ω = 0. Hence, capacitor C amounts to an open circuit.

 

Average power transferred to the resistor = 788.44 W

Average power transferred to the capacitor = 0 W

Total power absorbed by the circuit = 788.44 W

Inductance of inductor, L = 80 mH = 80 × 10−3 H

Capacitance of capacitor, C = 60 μF = 60 × 10−6 F

Resistance of resistor, R = 15 Ω

Potential of voltage supply, V = 230 V

Frequency of signal, ν = 50 Hz

Angular frequency of signal, ω = 2πν= 2π × (50) = 100π rad/s

The elements are connected in series to each other. Hence, impedance of the circuit is given as:

 

Current flowing in the circuit,

Average power transferred to resistance is given as:

PR= I2R

= (7.25)2 × 15 = 788.44 W

Average power transferred to capacitor, PC = Average power transferred to inductor, PL = 0

Total power absorbed by the circuit:

= PR + PC + PL

= 788.44 + 0 + 0 = 788.44 W

Hence, the total power absorbed by the circuit is 788.44 W.

 

Inductance, L = 0.12 H

Capacitance, C = 480 nF = 480 × 10−9 F

Resistance, R = 23 Ω

Supply voltage, V = 230 V

Peak voltage is given as:

V0 =

(a) Current flowing in the circuit is given by the relation,

Where,

I0 = maximum at resonance

At resonance, we have

Where,

ωR = Resonance angular frequency

∴Resonant frequency,

And, maximum current

(b) Maximum average power absorbed by the circuit is given as:

Hence, resonant frequency is

(c) The power transferred to the circuit is half the power at resonant frequency.

Frequencies at which power transferred is half, =

Where,

Hence, change in frequency,

∴

And,

Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half.

At these frequencies, current amplitude can be given as:

(d) Q-factor of the given circuit can be obtained using the relation,

Hence, the Q-factor of the given circuit is 21.74.

 

Inductance, L = 3.0 H

Capacitance, C = 27 μF = 27 × 10−6 F

Resistance, R = 7.4 Ω

At resonance, angular frequency of the source for the given LCR series circuit is given as:

Q-factor of the series:

To improve the sharpness of the resonance by reducing its ‘full width at half maximum’ by a factor of 2 without changing, we need to reduce R to half i.e.,

Resistance =

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

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

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

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

(d) If an iron core is inserted in the choke coil (which is in series with a lamp connected to the ac line), then the lamp will glow dimly. This is because the choke coil and the iron core increase the impedance of the circuit.

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

 

Input voltage, V1 = 2300

Number of turns in primary coil, n1 = 4000

Output voltage, V2 = 230 V

Number of turns in secondary coil = n2

Voltage is related to the number of turns as:

Hence, there are 400 turns in the second winding.

 

Total electric power required, P = 800 kW = 800 × 103 W

Supply voltage, V = 220 V

Voltage at which electric plant is generating power, V’ = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wire lines carrying power = 0.5 Ω/km

Total resistance of the wires, R = (15 + 15)0.5 = 15 Ω

A step-down transformer of rating 4000 − 220 V is used in the sub-station.

Input voltage, V1 = 4000 V

Output voltage, V2 = 220 V

Rms current in the wire lines is given as:

(a) Line power loss = I2R

= (200)2 × 15

= 600 × 103 W

= 600 kW

(b) Assuming that the power loss is negligible due to the leakage of the current:

Total power supplied by the plant = 800 kW + 600 kW

= 1400 kW

(c) Voltage drop in the power line = IR = 200 × 15 = 3000 V

Hence, total voltage transmitted from the plant = 3000 + 4000

= 7000 V

Also, the power generated is 440 V.

Hence, the rating of the step-up transformer situated at the power plant is 440 V − 7000 V.

 

The rating of a step-down transformer is 40000 V−220 V.

Input voltage, V1 = 40000 V

Output voltage, V2 = 220 V

Total electric power required, P = 800 kW = 800 × 103 W

Source potential, V = 220 V

Voltage at which the electric plant generates power, V’ = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wire lines carrying power = 0.5 Ω/km

Total resistance of the wire lines, R = (15 + 15)0.5 = 15 Ω

P = V1I

Rms current in the wire line is given as:

(a) Line power loss = I2R

= (20)2 × 15

= 6 kW

(b) Assuming that the power loss is negligible due to the leakage of current.

Hence, power supplied by the plant = 800 kW + 6kW = 806 kW

(c) Voltage drop in the power line = IR = 20 × 15 = 300 V

Hence, voltage that is transmitted by the power plant

= 300 + 40000 = 40300 V

The power is being generated in the plant at 440 V.

Hence, the rating of the step-up transformer needed at the plant is

440 V − 40300 V.

Hence, power loss during transmission =

In the previous exercise, the power loss due to the same reason is. Since the power loss is less for a high voltage transmission, high voltage transmissions are preferred for this purpose.