250+ TOP MCQs on AC Voltage Applied to a Resistor | Class12 Physics

Physics Aptitude Test for IIT JEE Exam on “AC Voltage Applied to a Resistor”.

1. What is the rms value of the current, if its instantaneous current value is 8 sin314 t?
a) 2√4 A
b) 10 A
c) 4√2 A
d) 50 A
Answer: c
Clarification: I = I₀ sinωt.
I_rms = (frac {I_0}{sqrt {2}})
I_rms = (frac {8}{(sqrt {2})})
I_rms = 4√2 A

2. What will be the rms value of the voltage, if the sinusoidal value voltage is given as E = 100 sin 314 t applied across a resistor of resistance 15 ohms?
a) 200 V
b) 70.71 V
c) 100 V
d) 33.87 V
Answer: b
Clarification: E = E₀ sinωt.
E_rms = (frac {E_0}{sqrt {2}})
E_rms = (frac {100}{sqrt {2}})
E_rms = 70.71 V

3. Determine the rms value of current in the circuit if a 50 Ω resistor is connected across an ac source of 200 V, 50 Hz?
a) 4 A
b) 200 A
c) 50 A
d) 1 A
Answer: a
Clarification: I_rms = (frac {E_{rms}}{R})
I_rms = (frac {200}{50})
I_rms = 4 A

4. Calculate the rms voltage, if the peak value of an a.c. supply is 300 V.
a) 561.4 V
b) 969.2 V
c) 625.1 V
d) 212.1 V
Answer: d
Clarification: E₀ = 300 V
E_rms = 0.707 E₀
E_rms = 0.707 × 300
E_rms = 212.1 V

5. Find out the net power consumed over a full cycle, if a 150 Ω resistor is connected to a 350 V, 100 Hz ac supply.
a) 2.3 W
b) 350 W
c) 805 W
d) 500 W
Answer: c
Clarification: I_rms = (frac {E_{rms}}{R}).
I_rms = (frac {350}{150})
I_rms = 2.3 A.
P_av = E_rms I_rms
P_av = 350 × 2.3
P_av = 805 W

6. When a sinusoidal voltage E = 200 sin 314 t is applied to a resistor of 10 Ω resistance, the power dissipated is 250 Watts.
a) True
b) False