Question

A 0.180-H inductor is connected in series with a 90.0 Ω resistor and an ac source. The voltage across the inductor is V_L = -(12.0 V) sin [(480 rad/s) t]. (a) Derive an expression for the voltage V_R across the resistor. (b) What is V_R at t = 2.00 ms?

Short answer

The voltage drop across resistor at t = 2 milliseconds is around 7.17 Volts

Step-by-step solution

Concept. 

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).

Given data

The inductance of the coil, L = 0.180 H

Series Resistance, R = 90.0 Ω

Expression of VR as a function of time

Voltage across an inductor is given by

$v_L=-I\omega Lsin\left(\omega t\right)$

Voltage equation given is

$v_L=-\left(12V\right)sin\left(480t\right)$

On comparing both equation we get,

$I\omega L=12,and\omega =480rad/s$

$$\begin{gathered} l=\frac{12.0V}{\omega L} \\ =\frac{12V}{\left(480rad/s\right)*\left(0.18H\right)} \\ =0.1389A \end{gathered}$$

Now, 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

$$\begin{gathered} v_R=IR\cos \left(\omega t\right) \\ =\left(0.1389A\right)\left(90\Omega \right)\cos \left(480t\right) \\ =\left(12.5V\right)\cos \left[\left(480rad/s\right)t\right] \end{gathered}$$

VR @ t = 2 milliseconds

Put t = 2 ms in the expression of v_R derived in previous step,

$$\begin{gathered} v_R=\left(12.5V\right)\cos \left[\left(480rad/s\right)t\right] \\ =\left(12.5V\right)\cos \left[\left(480rad/s\right)\left(2*{10}^{-3}s\right)\right] \\ =7.17V \end{gathered}$$

Therefore, the voltage drop across resistor at t = 2 milliseconds is around 7.17 Volts.