Question

A series RLC circuit with a $100\Omega$resistor dissipates $80W$when attached to a $120V/60Hz$power line. What is the power factor?

Short answer

The power factor of RLC circuit is$0.556$.

Step-by-step solution

Given Information

We are given that the resistor $R=100\Omega$, dissipating power $P=80W$and the voltage $V=120V$.

We have to find out the power factor of the RLC circuit.

Formula Used

Now, using the formula,

$P=\frac{V^2}{R}\times P_f$let this be equation 1

where P is power, V is voltage , R is resistor and $P_f$is power factor.

Simplify

Now put the values of all variables in equation 1,

We get $80=\frac{{\left(120\right)}^2}{100}{P}_f$

On rearranging,

We get, $P_f=\frac{80\times 100}{{\left(120\right)}^2}$

On solving,

$P_f=0.556$

The power factor of RLC circuit is$0.556$.