Question

An alternating current I = 14 sin (100 πt) A passes through a series combination of a resistor of 30 Ω and an inductor of `(2/(5pi))` H. Taking `sqrt2` = 1.4 calculate the power factor of the circuit.

Answer 1

Given: I = 14 sin (100 πt) A, R = 30 Ω, L = `2/(5pi)`H

From here

ω = 100π

`X_L = omegaL` (inductive reactance)

= `100pi xx 2/(5pi) = 40Omega`

Impedance, Z = `sqrt(R^2 + X_L^2) = sqrt((30)^2 + (40)^2)`

Z = `sqrt(900 + 1600) = 50Omega`

Power factor = `cosphi`

phasor diagram for finding `phi`,

`cosphi = R/Z`

= `30/50`

`cosphi = 3/5`

Power factor = 0.6