Question

Comprehension :

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

$\left(i\right)$ Find the expression of circuit current-

• A.
  $0.2\sin \left(1000t\right)\text{mA}$
• B.
  $2\sin \left(100t\right)\text{mA}$
• C.
  $2\sin \left(1000t\right)\text{mA}$
• D.
  $0.2\sin \left(100t\right)\text{mA}$

Answer 1

The correct option is A

$0.2\sin \left(1000t\right)\text{mA}$
Given, $V_R=2.0\sin \left({10}^{3}t\right)\text{V};R=100\Omega L=4\text{H};\omega ={10}^3$

Current through the resistor will be,

$I=I_0\sin \left({10}^{3}t\right)$

Where, $I_0=\frac{V_0}{R}=\frac{2}{100}=2\times {10}^{-2}$

$\therefore I=2\times {10}^{-2}\sin \left({10}^{3}t\right)\text{A}\text{or}0.2\sin \left(1000t\right)\text{mA}$

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