Bihar Board Class 12th Physics Solutions Chapter 7 Alternating Current
Bihar Board Class 12 Physics Alternating Current Textbook Questions and Answers
Question 1.
A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?
Answer:
Here, R = resistance = 100 Ω.
r.m.s. voltage, Ev = 220 V = Erms.
Frequency of A.C. supply, v = 50 Hz.
(a) Let Iv be the r.m.s. value of current in the circuit = Irms =?
Using the relation,
(b) We know that the power dissipated in an a.c. circuit is given by
P = Ev Iv cos Φ.
where Φ is the phase difference between current and voltage.
In a circuit containing resistor only, Φ = 0, thus cos Φ = cos 0 = 1.
∴ P = Ev. Iv.
= 220 x 2.2 = 484 W.
Question 2.
(a) The peak voltage of an ac supply is 300 V. What is the r.m.s. voltage?
(b) The r.m.s. value of current in an ac circuit is 10 A. What is the peak current?
Answer:
Here, peak value of a.c. supply, E0 = 300 V.
Ir.m.s = r.m.s. value of cttrrent = 10 A.
(a) Let Er m s be the r.m.s. value of the voltage =?
∴ Using the relation, 
(b) Let I0 be the peak value of the current.
we know that
∴ I0 =
Question 3.
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Answer:
Here, L = 44 mH = 44 x 10-3 H.
v = frequency of A.C. supply = 40 eps.
Er.m.s = r.m.s. value of A.C. voltage = 220 V.
Let Ir.m.s = r.m.s. value of current in the circuit =?
We know that Ir.m.s through an inductor circuit is given by
Where XL = ωL = 2πvL is the reactance of the inductor.
= 2π x 50 x 44 x 10-3 = 13.82 Ω.
∴ From (1),
Question 4.
A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Answer:
Here, Capacitance of the capacitor, C = 60 μF = 60 x 10-6 F. Frequency of A.C. supply, v = 60 Hz
r.m.s. value of A.C.voItage, Er.m.s = 110 V.
Let Ir.m.s be the r.m.s. value of current in the circuit =?
We know that Ir.m.s through a circuit having capacitor is given by,
Question 5.
In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Answer:
The net power absorbed by each circuit in Question 7.3. and 7.4 is
zero.
Explanation- We know that the power absorbed in an a.c. circuit is given by
Pav = Ev. Iv Cos Φ …………………………… (i)
where cos Φ is called power factor, being the angle between current and voltage. For a pure inductor and pure capacitor circuit,
∴
Thus from equation (1), Pav = 0 for each case.
Question 6.
Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?
Answer:
Here, L = 2.0 H, C = 32 μF = 32 x 10-6 F, R = 10 Ω, ωr, = resonant angular frequency =?
Q-factor of this circuit
We know that the resonant angular frequency is given by
The Q-value of this circuit is given by
Question 7.
A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Answer:
Here, C = 30 μF = 30 x 10-6 F
L = 27 mH = 27 x 10-3 H.
ω0 = angular frequency of oscillation =?
ω0 is given by,
Question 8.
Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
Answer:
Here, Qi = initial charge on the capacitor = 6 mC 6 x 10-3 C.
L = 27mH = 27 x 10-3 H
C = 30 μF = 3O x 10-6 F
Let V1 be the total energy stored in the circuit initially. It will be stored across the capacitor and is given by
As there is no resistance in the circuit given here, thus there is no damping of oscillation, hence the total energy will remain conserved during the LC oscillations.
Thus total energy at the later time = 0.6 J.
Question 9.
A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?
Answer:
Here, R = 20 Ω, L = 1.5 H, C = 35 μF = 35 x 10-6 F.
E. m = r.m.s. value of a.c. supply = 200 V.
When the frequency of the supply equals the natural frequency of the circuit, resonance takes place and the impedance of the circuit (Z) becomes equal to the resistance R i.e., the LCR circuit is purely resistive,
∴ Z = R = 20Ω.
Since the LCR circuit is resistive, so the phase angle between the current and voltage is zero.
i.e., Φ = 0°, and
Let P = average power transferred per cycle.
∴ P = Irms Erms cos 0°
= 10 x 200
= 2000 W.
= 2.0 KW.
Question 10.
A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 pH, what must be the range of its variable capacitor?
[Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radio wave.]
Answer:
Here, v1 = 800 kHz = 8 x 105 Hz, vz = 1200 kHz = 12 x 105 Hz.
L = 200 μH = 200 x 10-6 H = 2 x 10-4 H.
Let C1 and C2 be the capacity range between C1 and C2 =?
Using the relation,
Question 11.
The figure here shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 μF, R = 40 Ω.
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
Answer:
Here, L = 5.0 H, C = 80 μH = 80 x 10-6 F, R = 40 Ω, Er.m.s = 230 V.
(a) If ω0 be the resonant angular frequency = source frequency at resonance, then
If v0 = source frequency at resonance, then
(b) Let Z be the impedence of the circuit at resonance and I0 be the amplitude of current = peak value of current, then
