A circuit draws a power of 550 W from a 220 V – 50 Hz source. The power factor of the circuit is 0.8. A current in the circuit lags behind the voltage. Show that a capacitor of about $\frac{1}{42 π} \times 10^{- 2} F$ will have to be connected in the circuit to bring its power factor to unity.

Solution

Given,
Power drawn by the circuit, P = 550 W
Effective voltage, V_eff= 220 V
Frequency of a.c. source supply, f = 50 Hz
Power factor of circit, cos

ϕ

= 0.8

Power in the circuit is given by,

P = I_{V} E_{V} \cos ϕ

Therefore,

I_{V} = \frac{P}{E_{V} \cos ϕ} = \frac{550}{220 \times 0.8} A = 3.125 A

Resistance of the circuit,

R = \frac{P}{I_{V}^{2}} = \frac{550}{(3.125)^{2}} = 56.3 Ω

Now, using

\tan ϕ = \frac{ωL}{R}, we get

ωL = 42 Ω

Again,

ωL = \frac{1}{ωC}

[For power factor one]

C = \frac{1}{ω (ωL)}

= \frac{1}{100 π \times 42} F

= \frac{1}{42 π} \times 10^{- 2} F .

For Daily Free Study Material Join wiredfaculty