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Find the total voltage applied in a series RLC circuit when i=3mA, VL=30V, VC=18V and R=1000 ohms.
Explanation: Total voltage= VR+VL+VC.
VR=1000x3x103=3V.
Therefore, total voltage = 30+18+3=51V.
The power factor of an RL circuit 1/root2. If the frequency of a.c. is doubled, what will be the power factor?
When an emf E = 7cos wt is applied across a circuit, the current is I = 5coswt. What is the power factor for the circuit?
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
P=I_{rms} V_{rms} cos ϕ
Where, ϕ=90
cosϕ=0
therefore,P=0
hence the correct answer is option A.
If the instantaneous current in a circuit is given by i = 2 cos (ωt  φ) A, the rms value of the current is
i=2cost
=Imcost
So, Im=2amp
IRMS=Im/√2
=2/√2
=2amp
Find the true power given apparent power = 10 W and power factor = 0.5
The formula of true power is,
True power=Apparent power x power factor
So, true power=10 x 0.5
True power=5W
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_{m}v_{m} (cos ωt sin ωt)
pc=(i_{m}v_{m}/2)sin2ωt
We know that sin ωt = 0
Therefore, average power = 0
Cos x (power factor) = R/Z
= 10/20
=1/2
x= 60
In a circuit, the reactance X and resistance R are equal. What is the power factor?
Given that,
Reactance (X) =Resistance (R)
tanϕ = X/R
So, we have ϕ = π/4
Power factor = cosϕ = 1/√2
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