A 250 Ω resistor is connected in series with a 4.80 µF capacitor and an ac source. The voltage across the capacitor is v_C = (7.60 V) sin [(120 rad/s) t]. (a) Determine the capacitive reactance of the capacitor. (b) Derive an expression for the voltage v_R across the resistor.

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it. When an inductor is attached to an AC supply, the resistance produced by it is called inductive reactance (X_L).

Resistance is measure of opposition to the flow of current in a closed electrical circuit. It is measured in Ohm (Ω).

Series Resistance, R = 250 Ω

The capacitance of capacitor, C = 4.80 µF

Voltage across a capacitor is given by

$v_C=\frac{I}{\omega C}sin\left(\omega t\right)$

Voltage equation given is

$v_C=(7.6V)sin\left(120t\right)$

On comparing both equation we get,

localid="1664169758042" $\frac{I}{\omega C}=7.6,and\omega =120rad/s$

localid="1664169761543" $I=(7.6V)\omega C=(7.6V)(120rad/s)(4.80*{10}^{-6}F)=4.378*{10}^{-3}A$

We know that, localid="1664169766970" $V=I{X}_C$

So, the capacitive reactance is given by

localid="1664169770785" $X_C=\frac{V}{I}=\frac{7.6V}{4.378*{10}^{-3}A}=1736\Omega$

Therefore, the capacitive reactance of the capacitor is 1736 Ω

Expression of voltage across resistor is given by

$v_R=V_R\cos \left(wt\right)$

It can be manipulated using V_R = IR as

Therefore, the expression of voltage drop across resistor as a function of time is given by$v_R=(1.10V)cos\left[\right(120rad/s\left)t\right]$