Courses

# Test: AC Applied Across Resistor

## 10 Questions MCQ Test Physics Class 12 | Test: AC Applied Across Resistor

Description
This mock test of Test: AC Applied Across Resistor for JEE helps you for every JEE entrance exam. This contains 10 Multiple Choice Questions for JEE Test: AC Applied Across Resistor (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: AC Applied Across Resistor quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: AC Applied Across Resistor exercise for a better result in the exam. You can find other Test: AC Applied Across Resistor extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### Virtual value or effective value of a.c. is​

Solution:

Answer :- d

Solution :- as we know that H= (I0R)/2 * T/2 -----------------------(1)

If I(rms) be the rms value of ac then

H = I2(rms)R T/2 ---------------------(2)

From eq (1)&(2)

I2(rms)R T/2 = (I02R)/2 * T/2

I2(rms) = (I02)/2

= I(rms)= (I0)/(2)½

= I(rms) = 0.707 I0

QUESTION: 2

### Given the instantaneous value of current from a.c. source is I = 8 sin 623t. Find the r.m.s value of current​

Solution:

Compare the given eqn. with the standard from I=I0sinωt
I0=8, Irms=I0/√2=8/√2=5.656A

QUESTION: 3

### What is time constant

Solution:

Time constant is a measure of delay in an electrical circut resulting from either an inductor and resistor or capacitor and resistor. I will discuss rhe most common case which is resistor and capacitor, however the inductor resistor combination behaves in a similar manner. The time constant is equal to the value of the resistance in ohms multiplied by the value of capacitance in Farads. The time constant is measured in seconds . It represents the time for the voltage to decay to 1/2.72.

QUESTION: 4

Find the instantaneous voltage for an a.c. supply of 200V and 75 hertz​

Solution:

Answer :- b

Solution :-  f = 75hz

w=2πf

= 2 * π * 75

= 150π

E(max) = (2)^½ E(rms)

E(max) = 1.414 * 200

= 282.8V

E(ins) = E(max)sinwt

E(ins) = 282.8 sin 150πt

QUESTION: 5

If a capacitor of capacitance 9.2F has a voltage of 22.5V across it. Calculate the energy of the capacitor.

Solution:

We know that,
ω=(1/2)CV2
After putting the values,
=(1/2)x9.2x22.5x22.5
=2328.75J
Hence option B is the answer.

QUESTION: 6

Alternating current is represented by

Solution:

Alternating current is an electric current which periodically reverses direction, as opposed to direct current which flows only in one direction. And it can be easily represented by the periodic function.
So, I = Io sin wt or I = lo cos wt.

QUESTION: 7

What is the relationship between Em and E0

Solution:

peak value Em=2E0
so 2/π value is 0.637
therefore,
Em=-0.637 E0

QUESTION: 8

The only component that dissipates energy in ac circuit is:

Solution:

The only component that dissipates energy in ac circuit is the resistor because  Pure Inductive and pure capacitive circuits have no power loss.

QUESTION: 9

Which is more dangerous?​

Solution:

220 volt a.c. means the effective or virtual value of a.c. is 220 volt, i.e., Ev=220
As peak value E0=√2Ev
∴E0=1.414×220=311 volt
But 220 volt d.c. has the same peak value (i.e., 220 volt only).
Moreover, the shock of a.c. is attractive and that of d.c. is repulsive.
Hence 220 volt a.c. is more dangerous than 220 volt d.c.

QUESTION: 10

The average power dissipation in pure resistive circuit is:

Solution:

Power=IV
Where I=Vrms value of current=IV
And V=Vrms value of voltage=EV
Therefore, P=EVIV