Question

Comprehension :

A $100\Omega$ resistance is connected in series with a $4\text{H}$ inductor. The voltage across the resistor is, $V_R=2.0\sin \left({10}^{3}t\right)\text{V}.$

$\left(iii\right)$ Find amplitude of the voltage across the inductor.

• A.
  $40\text{V}$
• B.
  $60\text{V}$
• C.
  $80\text{V}$
• D.
  $90\text{V}$

Answer 1

The correct option is C

$80\text{V}$
The potential drop across the inductor is,

$V_L=I_0{X}_L$

$\because I_0=2\times {10}^{-2}\text{A};X_L=4\times {10}^3\Omega$

$\therefore V_L=80\text{V}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.