Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 1 (Q.No. 46)
46.
In a circuit of figure, Vs = 10 cos(ωt) power drawn by the 2Ω resistor is 4 watts. The power factor is
Answer: Option
Explanation:
Vs = 10 cos ωt,
Let I be the current, then P = I2 R
I is the rms current.
Prms = 
I2 R 4 I2 = 4 I = 2A.
Total power drawn = Vm Im cos θ
10 x 2 cos θ = 10 cos θ
Total power cannot be determine until value of R = 12 Ω is given.
Discussion:
4 comments Page 1 of 1.
KRISHNA said:
1 year ago
@All.
Here is my solution;
Given
Vs = 10 cos(wt)
Vm = 10
Vrms = Vm/√2 = 10/√2 = 5√2.
Given
P = 4 W, R = 2.
P = I^2 * R.
Irms=√2.
Power factor = real power/apparent power
=4/Vrms *Irms.
=4/10= 0.4.
Here is my solution;
Given
Vs = 10 cos(wt)
Vm = 10
Vrms = Vm/√2 = 10/√2 = 5√2.
Given
P = 4 W, R = 2.
P = I^2 * R.
Irms=√2.
Power factor = real power/apparent power
=4/Vrms *Irms.
=4/10= 0.4.
Gia said:
10 years ago
Shouldn't it be 0.4 since real power = VIcosθ = 10cosθ = 4W? so cosθ = 4/10 = 0.4.
Vasanthakumara D said:
6 years ago
Sir, how to calculate the answer please explain sir?
Sid said:
8 years ago
0.4 is the correct answer.
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