NCERT MOST IMPORTANT QUESTIONS CLASS – 12 | PHYSICS IMPORTANT QUESTIONS | CHAPTER – 7 | ALTERNATING CURRENT | EDUGROWN |

In This Post we are  providing Chapter- 7 ALTERNATING CURRENT NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ALTERNATING CURRENT

Question 1.
The figure shows how the reactance of an Inductor varies with frequency. (a) Calculate the value of the Inductance of the Inductor using Information given In the graph. (b) If this Inductor is connected In senes to a resistor of 8 ohms, find what would be the impedance at 300 Hz.

Answer:
(a) We know that X_L = 2πfL or L = XL2πf.
Now slope of the graph is
XLf=8−6400−300=2100 = 0.02

Therefore L is L = XL2πf=0.022×3.14 = 0.0032 H

(b) NowR = 8 ohm, f = 300 Hz, Z = ?
Now Z = R2+X2L−−−−−−−√ . Therefore we have
Z = R2+X2L−−−−−−−√ = (8)2+(6)2−−−−−−−−−√ = 10 Ω

Question 2.
A 25.0 μF capacitor, a 0.10-henry Inductor, and a 25.0-ohm resistor are connected in series with an ac source whose emf is E= 310 s. In 314 t (i) what is the frequency of the .mf? (ii) Calculate (a) the reactance of the circuit (b) the Impedance of the circuit and (C) the current in the circuit.
Answer:
Given C = 25.0 μF, L = 0.10 henry, R = 25.0 ohm, E_o = 310V,
Comparing with the equation E = E_o sin ωt,
we have
(i) ω = 314 or f = 50 Hz

Question 3.
A sinusoidal voltage V = 200 sin 314 t Is applied to a resistor of 10 ohms. Calculate (i) rms value of current (ii) rms value of voltage and (iii) power dissipated as heat in watt.
Answer:
V_o = 200 V, ω = 314 rads^-1 , V_rms = ?, l_rms = ?, P = ?

Question 4.
Find the inductance of the inductor used in series with a bulb of resistance 10 ohms connected to an ac source of 80 V, 50 Hz. The power factor of the circuit is 0.5. Also, calculate the power dissipation in the circuit.
Answer:
Given R = 10 ohm, V = 80 Hz, f = 50 Hz, cos Φ = 0.5 , L = ?, P = ?
Using the formula
cos Φ = RR2+X2L√
or
0.25(R² + X²) = R²
Or
O.25X_L² = O.75R²
or
X_L = 17.32 ohm

Now using X_L = 2πfL we have
L = XL2πf=17.322×3.14×50 = 0.055 H

Now Z = R2+X2L−−−−−−−√
= (10)2+(17.32)2−−−−−−−−−−−−−√ = 20 Ω

Now P_av = l_rms V_rms cos Φ = V2mZ × cos Φ
or
Pav = (80)22×20 × 0.5 = 160 W

Question 5.
When an alternating voltage of 220 V is applied across a device X, a current of 0.5 A flows through the circuit and Is in phase with the applied voltage. When the same voltage is applied across a device Y, the same current again flows through it, but it leads the voltage by π/2. If element ‘X’ is a pure resistor of 100 ohms,
(a) name the circuit element ‘Y’ and
Answer:
The element Y is a capacitor.

(b) calculate the rms value of current, if rms value of voltage is 141 V.
Answer:
The value of Xc is obtained as below

X_C = VI=2200.5 = 440 ohm

Therefore impedance of the circuit
Z = R2+X2C−−−−−−−√=(100)2+(440)2−−−−−−−−−−−−√ = 451.2 ohm

Therefore rms value of current V 141
l = VrmsZ=141451.2 = 0.3125 A

Question 6.
When an alternating voltage of 220 V is applied across a device X, a current of 0. 5 A flows through the circuit and is in phase with the applied voltage. When the same voltage is applied across a device V, the same current again flows through it, but it lags the voltage by π/2.
(a) Name the devices X and Y.
Answer:
The element X is a resistor and Y is an inductor.

(b) Calculate the current flowing through the circuit when the same voltage is applied across the series combination of the two devices X and Y.
Answer:
Now both R and XL are the same and are given by
R = X_L = 2200.5 = 440 ohm

Hence impedance of the circuit
Z = R2+X2L−−−−−−−√
= (440)2+(440)2−−−−−−−−−−−−√
= 622.2 ohm

Therefore current flowing through the circuit is
l = VZ = 220622.2 = 0.353 A

Question 7.
A 15.0 μF capacitor Is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) In the circuit. If the frequency Is doubled, what happens to the capacitive reactance and the current?
Answer:
Given C= 15.0 μF= 15 × 10^-6 F, V= 220 V,
f = 50 HZ, X_C = ? l_m = ?

The capacitive reactance is

Now l_rms = VrmsXc=220212 = 1.04 A

Peak value of current
l_m = 2–√ × lrms = 1.4.1 × 1.04 = 1.47 A

This current oscillates between + 1.47A and – 1.47 A, and is ahead of the voltage by π/2.

If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

Question 8.
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 μF. Find
(a) the impedance of the circuit;
(b) the phase difference between the voltage across the source and the current;
(c) the power dissipated in the circuit; and
(d) the power factor.
Answer:
Given V_m = 283 V, f = 50 Hz, R = 3 Ω
L = 25.48 mH, and C = 796 μF.
(a) To find the impedance of the circuit, we first calculate XL and Xc
X_L = 2πfL = 2 × 3.14 × 50 × 25.48 × 10^-3 = 8 Ω
X_C = 12πfC
= 12×3.14×50×796×10−6 = 4 Ω

Therefore impedance of the circuit is
Z = R2+(XL−XC)2−−−−−−−−−−−−−−√
= 32+(8−4)2−−−−−−−−−−√
= 5 Ω

(b) Phase difference
Φ = tan^-1XL−XcR = tan-1 8−43 = 53.1°
Since Φ is positive therefore voltage leads current by the above phase.

(c) The power dissipated in the circuit is
P = V2rmsZ=V2m2Z=(283)22×5 = 8008.9 W
(d) power factor = cos Φ = cos 53.1° = 0.6

Question 9.
A capacitor and a resistor are connected in series with an ac source. If the potential difference across the C, R is 120 V and 90 V respectively, and if rms current of the circuit is 3 A, calculate the (i) impedance and (ii) power factor of the circuit.
Answer:
Given V_C = 120 V, V_R = 90 V, f = 3 A. and R = 90/3 = 30 ohm ,

Effective voltage in the circuit
V = V2C+V2R−−−−−−−√=(120)2+(90)2−−−−−−−−−−−√= 150 V .
(i) Therefore impedance of the circuit
Z = Vl=1503 = 50 Ω.

(ii) Now power factor of the circuit is
cos Φ = RZ=3050 = 0.6

Question 10.
An inductor 200 mH, a capacitor C, and a resistor 10 ohm are connected in series with 100 V, 50 Hz ac source. If the current and the voltage are in phase with each other, calculate the capacitance of the capacitor.
Answer:
When current and voltage are in phase then X_L = X_C