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Question

In a series LR circuit, power of 400W is dissipated from a source of 250V,50Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as n3πμF, then the value of n is __________.


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Solution

Step 1: Given data

Power, P=400W

Source voltage,Vrms=250V

Frequency, f=50Hz

Power factor,cosϕ=0.8

Step 2: To find

We have to find the value of n in the equation C=n3πμF

Step 3: Finding Impedance, Z

We know, power is given by,

P=VrmsIrmscosϕP=VrmsVrmsZcosϕ(Irms=VrmsZ)Z=Vrms2Pcosϕ

Where Irms is rms current and Z is the impedance of the circuit.

Substituting the values,

Z=2502400×0.8Z=125Ω

Step 4: Finding resistance(R), inductive reactance(XL) and capacitive reactance(XC)

Also,

cosϕ=RZR=ZcosϕR=125×0.8R=100Ω

Impedance,

Z=R2+XL2XL=Z2-R2XL=1252-1002XL=1252-1002XL=75Ω

Since the power factor is unity, XL=XC=75Ω

Step 5: Finding capacitance C and hence n

XC=75Ω1Cω=7512πfC=75ω=2πf12π×50×C=75f=50HzC=12π×50×75C=4003πμF

Since, C=4003πμF, we get n=400

The value of n is 400


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