Question

In a parallel circuit, ET = 240 V, R = 5.6 kΩ, and XL = 3.6 kΩ. What is P?

Answer 1

Real power,P is 10.286 Watts

Explanation:

We are given that in a parallel circuit,
`E_(T)=240V`,R=5.6kΩ and
`X_(L)`=3.6kΩ

We are asked to find P which means real power, that is power due to resistance.

Let us find current through resistor first.

That is
`I_(R)=(E_(T))/(R)=(240)/(5.6)=(300)/(7)=42.86`milli Amps=0.04286 A

Hence
`P=E_(T)I_(R)`

=240X0.04286=10.286 Watts