A resistor of 6 Ω and an inductor having inductive reactance of 8 Ω are connected in series to a 250 V, 50 Hz supply. Calculate the active power consumed.

Question

Question

This question was previously asked in

SSC JE Electrical 15 Nov 2022 Shift 2 Official Paper

Answer 1

Concept

The active power in an RL circuit is given by:

P = VI cosϕ

where, V = Voltage

I = Current

cos ϕ = Power factor

The value of current is given by:

\(I={V\over \sqrt{(R)^2+(X_L)^2}}\)

The power factor is given by:

\(cos\space \phi=cos[tan^{-1}({X_L\over R})]\)

where, R = Resistance

XL = Inductive Reactance

Calculation

Given, R = 6 Ω and XL = 8 Ω

V = 250 volts

\(I={250\over \sqrt{(6)^2+(8)^2}}\)

I = 25 A

\(cos\space \phi=cos[tan^{-1}({8\over 6})]\)

cos ϕ = 0.6

P = 250 × 25 × 0.6

P = 3.75 kW