Class 12 Alternating Current CBSE Questions & Answers

Class 12 · Alternating Current

This is Physics Class 12 Alternating Current CBSE Questions & Answers. There are 15 questions in this test with each question having around four answer choices.

Questions & Answers

1
The current amplitude in a pure inductor in a radio receiver is to be 250 \(\mu {\rm{A}}\) when the voltage amplitude is 3.60 V at a frequency of 1.60 MHz (at the upper end of the AM broadcast band). If the voltage amplitude is kept constant, what will be the current amplitude through this inductor at 16.0 MHz?
  • A
    33.0 \(\mu {\rm{A}}\)
  • B
    35.0 \(\mu {\rm{A}}\)
  • C
    20.0 \(\mu {\rm{A}}\)
  • D
    25.0 \(\mu {\rm{A}}\)
    Correct
2
A resistor is connected in series with a capacitor. The voltage across the resistor is vR = (1.20 V) cos(2500 rad/s)t . Capacitive reactance is
  • A
    60 \(\Omega \)
  • B
    80 \(\Omega \)
    Correct
  • C
    70 \(\Omega \)
  • D
    90 \(\Omega \)
3
In a series RLC circuit R = 300 \(\Omega \) , L = 60 mH, C = 0.50 \(\mu {\rm{F}}\) applied voltage V= 50 V and \(\omega \) = 10,000 rad/s. Reactances \({{\rm{X}}_{\rm{L}}}\), \({{\rm{X}}_{\rm{C}}}\) and Z are
  • A
    450 \(\Omega \), 200 \(\Omega \) and 450 \(\Omega \)
  • B
    600 \(\Omega \), 200 \(\Omega \) and 500
    Correct
  • C
    550 \(\Omega \), 200 \(\Omega \) and 500 \(\Omega \)
  • D
    500 \(\Omega \), 250 \(\Omega \) and 500 \(\Omega \)
4
In a series RLC circuit R = 300 x \(\Omega \) , L = 60 mH, C = 0.50 \(\mu {\rm{F}}\) applied voltage V= 50 V and ω = 10,000 rad/s. voltages \({{\rm{V}}_{{\rm{R}},}}{{\rm{V}}_{\rm{L}}}\) and \({{\rm{V}}_{\rm{C}}}\) are
  • A
    36 V, 65 V, 20 V
  • B
    36 V, 66 V, 25 V
  • C
    35 V, 60 V, 26V
  • D
    30 V, 60 V, 20 V
    Correct
5
An electric hair dryer is rated at 1500 W (the average power) at 120 V (the rms voltage). Calculate (a) the resistance, (b) the rms current, and (c) the maximum instantaneous power. Assume that the dryer is a pure resistor.
  • A
    8.6 \(\Omega \), 11.5 A, 3000 W
  • B
    10.6 \(\Omega \) , 13.5 A, 3000 W
  • C
    7.6 \(\Omega \) , 14.5 A, 3000 W
  • D
    9.6 \(\Omega \), 12.5 A, 3000 W
    Correct
6
A hair dryer meant for 110V 60Hz is to be used in India . If 220 V is the supply voltage in India , the turns ratio for a transformer would be
  • A
    step-up 1:2
  • B
    step-down 3:1
  • C
    step-down 2.5:1
  • D
    step-down 2:1
    Correct
7
Mr Iyer takes a grinder rated 220 V 50 Hz to the U.S where supply is 110 60 Hz . He needs to use a
  • A
    step-up 1:3.6
  • B
    step-up 1:2.5
  • C
    step-up 1:2
    Correct
  • D
    step-up 1:3
8
You have a special light bulb with a very delicate wire filament. The wire will break if the current in it ever exceeds 1.50 A, even for an instant. What is the largest root-mean-square current you can run through this bulb?
  • A
    1.26 A
  • B
    1.46 A
  • C
    1.56 A
  • D
    1.06 A
    Correct
9
The voltage across the terminals of an ac power supply varies with time according to \({\rm{Vcos}}\omega {\rm{t}}\). The voltage amplitude is V = 45.0 V. Root-mean-square potential difference is
  • A
    35.8 V
  • B
    37.8 V
  • C
    31.8 V
    Correct
  • D
    33.8 V
10
The voltage across the terminals of an ac power supply varies with time according to \({\rm{Vcos}}\omega {\rm{t}}\). The voltage amplitude is V = 45.0 V. Average potential difference between the two terminals of the power supply is
  • A
    35.8 V
  • B
    37.8 V
  • C
    zero V
    Correct
  • D
    33.8 V
11
An inductor with L = 9.50 mH is connected across an ac source that has voltage amplitude 45.0 V. Phase angle for the source voltage relative to the current is
  • A
    90\(^\circ \)
    Correct
  • B
    -90\(^\circ \)
  • C
    180\(^\circ \)
  • D
    120\(^\circ \)
12
An inductor with L = 9.50 mH is connected across an ac source that has voltage amplitude 45.0 V. Frequency of the source that results in a current amplitude of 3.90 A is
  • A
    180 Hz
  • B
    150 Hz
  • C
    193 Hz
    Correct
  • D
    129 Hz
13
The wiring for a refrigerator contains a starter capacitor. A voltage of amplitude 170 V and frequency 60.0 Hz applied across the capacitor is to produce a current amplitude of 0.850 A through the capacitor. Capacitance required is
  • A
    15.3 \(\mu {\rm{F}}\)
  • B
    13.3 \(\mu {\rm{F}}\)
    Correct
  • C
    17.8 \(\mu {\rm{F}}\)
  • D
    23.4 \(\mu {\rm{F}}\)
14
Reactance of a 3.00-H inductor at a frequency of 80.0 Hz is
  • A
    1490 \(\Omega \)
  • B
    1510 \(\Omega \)
    Correct
  • C
    1480 \(\Omega \)
  • D
    1500 \(\Omega \)
15
Inductance of an inductor whose reactance is 120 \(\Omega \) at 80.0 Hz is
  • A
    0.199 H
  • B
    0.209 H
  • C
    0.219 H
  • D
    0.239 H
    Correct