Alternating Current Test

Result

Alternating Current Test - 5

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Virtual value or effective value of a.c. is

A
-0.637I₀

B
-0.707I₀

C
0.637I₀

D
0.707I₀

Solution

Answer :- d
Solution :- as we know that H= (I₀²R)/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
1 / -0

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

A
5.656 A

B
8 A

C
2.848 A

D
1.414 A

Solution

Compare the given eqn. with the standard from I=I0sinωt
I₀=8, Irms=I₀/√2=8/√2=5.656A
Question 3
1 / -0

What is time constant

A
characteristic time with which current grows and decays

B
the time current takes to reach the minimum value only

C
time for the complete cycle

D
the time current takes to reach the peak value only

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
1 / -0

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

A
E = 282.8 sin 50πt

B
E = 282.8 sin 150πt

C
E = 282.8 sin 75πt

D
E = 282.8 sin 100πt

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
1 / -0

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

A
5062.5 J

B
2328.75 J

C
50.625 J

D
50625 J

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
1 / -0

Alternating current is represented by

A

B

C

D

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 = I_o sin wt or I = l_o cos wt.
Question 7
1 / -0

What is the relationship between E_m and E₀

A
E_m=-0.537 E₀

B
E_m=-0.737 E₀

C
E_m=-0.637 E₀

D
E_m=-0.707 E₀

Solution

peak value E_m=2E₀/π
so 2/π value is 0.637
therefore,
E_m=-0.637 E₀
Question 8
1 / -0

The only component that dissipates energy in ac circuit is:

A
Capacitor

B
Inductor

C
Resistors

D
None of these

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
1 / -0

Which is more dangerous?

A
220 V a.c

B
220 V d.c

C
Both 220 V a.c. and 220 V d.c

D
Both 220 V a.c. and 220 V d.c. are not dangerous

Solution

220 volt a.c. means the effective or virtual value of a.c. is 220 volt, i.e., E_v=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
1 / -0

The average power dissipation in pure resistive circuit is:

A

B

C

D

Solution

Power=IV
Where I=V_rms value of current=I_V
And V=V_rms value of voltage=E_V
Therefore, P=E_VI_V