# Physics Questions and Answers – AC Voltage Applied to a Resistor

«
»

This set of Physics Aptitude Test for IIT JEE Exam focuses 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

Explanation: I = I0 sinωt.
Irms = $$\frac {I_0}{\sqrt {2}}$$
Irms = $$\frac {8}{(\sqrt {2})}$$
Irms = 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

Explanation: E = E0 sinωt.
Erms = $$\frac {E_0}{\sqrt {2}}$$
Erms = $$\frac {100}{\sqrt {2}}$$
Erms = 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

Explanation: Irms = $$\frac {E_{rms}}{R}$$
Irms = $$\frac {200}{50}$$
Irms = 4 A
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

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

Explanation: E0 = 300 V
Erms = 0.707 E0
Erms = 0.707 × 300
Erms = 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

Explanation: Irms = $$\frac {E_{rms}}{R}$$.
Irms = $$\frac {350}{150}$$
Irms = 2.3 A.
Pav = Erms Irms
Pav = 350 × 2.3
Pav = 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

Explanation: Yes, this is a true statement. The explanation is given as follows:
E = E0 sinωt.
Erms = $$\frac {E_0}{\sqrt {2}} = \frac {50}{\sqrt {2}}$$
Erms = 35.35 V.
Irms = $$\frac {E_{rms}}{R} \, = \, \frac {35.5}{5}$$
Irms = 7.07 A.
Power dissipated = Ev × Iv = 35.35 × 7.07
Power dissipated = 249.92 W ≈ 250 W.

7. Which value of the current do you measure with an a.c. ammeter?
a) Instantaneous current
b) Root mean square value
c) Electromotive force
d) Peak current

Explanation: Root mean square value of the current can be measured with the help of an a.c. ammeter. It is used to measure the alternating current that flows through any branch of an electric circuit is called AC ammeter.

8. Calculate the value of the peak current, if the rms value of current in an ac circuit is 50 A?
a) 70.7 A
b) 65 A
c) 50 A
d) 1 A

Explanation: Irms = 50 A.
I0 = √2 Irms
I0 = 1.414 × 50
I0 = 70.7 A
Therefore, the peak current is 70.7 A.

Sanfoundry Global Education & Learning Series – Physics – Class 12.