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# Test: AC Applied Across Resistor

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

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Test: AC Applied Across Resistor - Question 1

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

Detailed Solution for Test: AC Applied Across Resistor - Question 1

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

Test: AC Applied Across Resistor - 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​

Detailed Solution for Test: AC Applied Across Resistor - Question 2

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

Test: AC Applied Across Resistor - Question 3

### What is time constant

Detailed Solution for Test: AC Applied Across Resistor - Question 3

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.

Test: AC Applied Across Resistor - Question 4

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

Detailed Solution for Test: AC Applied Across Resistor - Question 4

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

Test: AC Applied Across Resistor - Question 5

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

Detailed Solution for Test: AC Applied Across Resistor - Question 5

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.

Test: AC Applied Across Resistor - Question 6

Alternating current is represented by

Detailed Solution for Test: AC Applied Across Resistor - Question 6

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.

Test: AC Applied Across Resistor - Question 7

What is the relationship between Em and E0

Detailed Solution for Test: AC Applied Across Resistor - Question 7

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

Test: AC Applied Across Resistor - Question 8

The only component that dissipates energy in ac circuit is:

Detailed Solution for Test: AC Applied Across Resistor - Question 8

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

Test: AC Applied Across Resistor - Question 9

Which is more dangerous?​

Detailed Solution for Test: AC Applied Across Resistor - Question 9

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.

Test: AC Applied Across Resistor - Question 10

The average power dissipation in pure resistive circuit is:

Detailed Solution for Test: AC Applied Across Resistor - Question 10

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

## Physics Class 12

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## Physics Class 12

157 videos|452 docs|213 tests

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