Page No 267: - Chapter 7 Alternating Current Additional Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 7.12:

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?

Answer:

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.

Question 7.13:

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 current maximum and the voltage maximum?

Answer:

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, V₀ = 

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.

Question 7.16:

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.

Answer:

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.

Question 7.17:

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.

Answer:

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:

Question 7.18:

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’.]

Answer:

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.

Question 7.19:

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.

Answer:

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.