Question

A series RLC circuit attached to a $120V/60Hz$power line draws a $2.4A$rms current with a power factor of $0.87$. What is the value of the resistor?

Short answer

The value of resistor is$43.5\Omega$.

Step-by-step solution

Given Information

We are given that rms current $I=2.4A$,Voltage is $V=120V$and power factor is $P_f=0.87$

and we have to find out the value of resistor.

Formula Used

Now, using the formula,

We get $P=\left(V\right)\left(I\right)\left(P_f\right)$

where V is voltage, I is current and $P_f$is the power factor.

Putting the value of variables in equation,

We get,

$P=\left(120\right)(2.4)(0.87)=250.56$W

Simplify

Now, using the formula

$P=I^{2}R$,

where P is power, I is current and R is resistor,

Putting all the variable in equation we get,

$250.56={(2.4)}^{2}R$

On solving,

we get

$R=43.5\Omega$

The value of resistor is$43.5\Omega$.