Alternating Current Test

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Alternating Current Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

Find the total voltage applied in a series RLC circuit when i=3mA, VL=30V, VC=18V and R=1000 ohms.

A
3.95V

B
32.67V

C
6.67V

D
51V

Solution

Explanation: Total voltage= VR+VL+VC.
VR=1000x3x10-3=3V.
Therefore, total voltage = 30+18+3=51V.
Question 2
1 / -0

The power factor of an RL circuit 1/root2. If the frequency of a.c. is doubled, what will be the power factor?

A
1/√7

B
1/√5

C
1/√3

D
1/√11

Solution
Question 3
1 / -0

When an emf E = 7cos wt is applied across a circuit, the current is I = 5coswt. What is the power factor for the circuit?

A
infinite

B
3/4

C
zero

D
1

Solution

Since E and I are in the same phase.
Therefore, phase difference will be 0 and since power factor= cosx (where x= phase difference) and x =0
therefore, cos x or power factor will be =1
Question 4
1 / -0

What is value of power factor if Ø = 90°

A
0

B
0.5

C
1

D
infinite

Solution

P=I_rms V_rms cos ϕ
Where, ϕ=90
cosϕ=0
therefore,P=0
hence the correct answer is option A.
Question 5
1 / -0

Power in an ac circuit is equal to

A
Instantaneous voltage X Instantaneous current

B
Instantaneous voltage X current at an instant

C
Voltage at an instant X Instantaneous current

D
Both b and c
Question 6
1 / -0

If the instantaneous current in a circuit is given by i = 2 cos (ωt - φ) A, the rms value of the current is

A
√2 A

B
2√2 A

C
2 A

D
0 A

Solution

i=2cost
=Imcost
So, Im=2amp
IRMS=Im/√2
=2/√2
=2amp
Question 7
1 / -0

Find the true power given apparent power = 10 W and power factor = 0.5

A
0.5 W

B
0.05 W

C
5 W

D
50 W

Solution

The formula of true power is,
True power=Apparent power x power factor
So, true power=10 x 0.5
True power=5W
Question 8
1 / -0

What is the average power/cycle in a capacitor?

A
0

B
infinite

C
E₀L₀ cos Ø

D
E₀L₀

Solution

The average power consumed/cycle in an ideal capacitor is 0.
The average power consumed in an ideal capacitor is given as based on the instantaneous power which is supplied to the capacitor:
p_c = iv= (i_m cos ωt)(v_m sin ωt)
p_c = i_mv_m (cos ωt sin ωt)
pc=(i_mv_m/2)sin2ωt
We know that sin ωt = 0
Therefore, average power = 0
Question 9
1 / -0

What is phase angle given R = 10, Z = 20

A
30°

B
45°

C
60°

D
90°

Solution

Cos x (power factor) = R/Z
= 10/20
=1/2
x= 60
Question 10
1 / -0

In a circuit, the reactance X and resistance R are equal. What is the power factor?

A
1/2

B
1/√2

C
√2

D
1

Solution

Given that,
Reactance (X) =Resistance (R)
tanϕ = X/R
So, we have ϕ = π/4
Power factor = cosϕ = 1/√2

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Re-Attempt Weekly Quiz Competition

Alternating Current Test - 9

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Alternating Current Test - 9

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SHARING IS CARING

Question 1

3.95V

32.67V

6.67V

51V

Solution

Question 2

1/√7

1/√5

1/√3

1/√11

Solution

Question 3

infinite

3/4

zero

1

Solution

Question 4

0

0.5

1

infinite

Solution

Question 5

Instantaneous voltage X Instantaneous current

Instantaneous voltage X current at an instant

Voltage at an instant X Instantaneous current

Both b and c

Question 6

√2 A

2√2 A

2 A

0 A

Solution

Question 7

0.5 W

0.05 W

5 W

50 W

Solution

Question 8

0

infinite

E0L0 cos Ø

E0L0

Solution

Question 9

30°

45°

60°

90°

Solution

Question 10

1/2

1/√2

√2

1

Solution

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E₀L₀ cos Ø

E₀L₀