Chapter 13 AC Circuits Solutions

Maharashtra State Board 12th Physics Solutions Chapter 13 AC Circuits

1. Choose the correct option.

i) If the RMS current in a 50 Hz AC circuit is 5A, the value of the current seconds after its value becomes zero is
(A) 5 A
(B) 5 A
(C) A
(D) A
Answer:
(B) 5 A

ii) A resistor of 500 Ω and an inductance of 0.5 H are in series with an AC source which is given by V = 100 sin (1000t). The power factor of the combination
(A)
(B)
(C) 0.5
(D) 0.6
Answer:
(A)

iii) In a circuit L, C & R are connected in series with an alternating voltage of frequency f. the current leads the voltage by 450. The value of C is

Answer:
(B)

iv) In an AC circuit, e and i are given by e = 150 sin (150t) V and i = 150 sin (150 t + ) A. the power dissipated in the circuit is
(A) 106W
(B) 150W
(C) 5625W
(D) Zero
Answer:
(C) 5625W

v) In a series LCR circuit the phase difference between the voltage and the current is 45°. Then the power factor will be
(A) 0.607
(B) 0.707
(C) 0.808
(D) 1
Answer:
(B) 0.707

2. Answer in brief.

i) An electric lamp is connected in series with a capacitor and an AC source is glowing with a certain brightness. How does the brightness of the lamp change on increasing the capacitance ?
Answer:
Impedance, Z = , where R is the resistance of the lamp, w is the angular frequency of AC and C is the capacitance of the capacitor connected in series with the AC source and the lamp. When C is increased, decreases. Hence, Z increases.
Power factor, cos Φ =
As Z increases, the power factor decreases.
Now, the average power over one cycle,
P_av= v_rmsi_rmscos Φ
= V_rms cos Φ
=
∴ P_avdecreases as Z increases and cos Φ decreases.
As the current through the lamp decreases, the brightness of the lamp will decrease when C is increased.

ii) The total impedance of a circuit decreases when a capacitor is added in series with L and R. Explain why ?
Answer:
For an LR circuit, the impedance,
Z_LR= , where X_Lis the reactance of the inductor.
When a capacitor of capacitance C is added in series with L and R, the impedance,
Z_LCR= because in the case of an inductor the current lags behind the voltage by a phase angle of rad while in the case of a capacitor the current leads the voltage by a phase angle of rad. The decrease in net reactance decreases the total impedance (Z_LCR< Z_LR).

iii) For very high frequency AC supply, a capacitor behaves like a pure conductor. Why ?
Answer:
The reactance of a capacitor is X_C= , where f is the frequency of the AC supply and C is the capacitance of the capacitor. For very high frequency, f, X_Cis very small. Hence, for very high frequency AC supply, a capacitor behaves like a pure conductor.

iv) What is wattless current ?
Answer:
The current that does not lead to energy consumption, hence zero power consumption, is called wattless current.
In the case of a purely inductive circuit or a purely capacitive circuit, average power consumed over a complete cycle is zero and hence the corresponding alternating current in the circuit is called wattless current.
[Note : In this case, the power factor is zero.]

v) What is the natural frequency of L C circuit ? What is the reactance of this circuit at this frequency
Answer:
The natural frequency of LC circuit is ,
where L is the inductance and C is the capacitance. The reactance of this circuit at this frequency is

(b) P_av= e_rmsi_rmscos Φ
The factor cos Φ is called as power factor. For circuits used for transporting electric power, a low power factor means the power available on transportation is much less than e_rmsi_rmsIt means there is significant loss of power during transportation.

(b)

(c) X_C= . Thus X_C∝ , where f is the frequency of AC. Suppose C = pF
For f= 100 Hz, X_C= 1 × 10⁷Ω = 10MΩ;
for f = 200 Hz, X_C= 5 MΩ;
for f = 300 Hz, X_C= MΩ;
for f = 400 Hz, X_C= 2.5 MΩ
for f = 500 Hz, X_C= 2 MΩ and so on

(d)

The phasor representing the peak emf (e₀) makes an angle (ωt) in an anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by 90°, the phasor representing the peak current (i₀) is turned 90° anticlockwise with respect to the phasor representing emf e₀. The projections of these phasors on the vertical axis give instantaneous values of e and i.

(2) In a purely resistive circuit, current and voltage are always in phase.
When reactance is not zero, there is nonzero phase difference between current and voltage.

(3) Resistance does not depend on the frequency of AC.
Reactance depends on the frequency of AC. In case of an inductor, reactance increases linearly with frequency. In case of a capacitor, reactance decreases as frequency of AC increases; it is inversely proportional to frequency.

(4) Resistance gives rise to production of Joule heat in a component.
In a circuit with pure reactance, there is no production of heat.

According to Kirchhoff’s voltage law, as the resistance of the inductor is assumed to be zero, we

where C is the constant of integration. C must be time independent and have the dimension of current. As e oscillates about zero, i also oscillates about zero and hence there cannot be any time independent component of current.
∴ C = 0. ∴ i = –cos ωt = – sin( – ωt)
∴ i = sin(ωt – ) ……………. (3)
as sin (-θ) = – sin θ
From Eq. (3), i_peak= i₀=
∴ i = i₀sin(ωt – ) ………………. (4)
Comparison of this equation with e = e₀sin ωt shows that e leads i by rad, i.e., the voltage is ahead of current by rad in phase.