Question

The figure shows a circuit that contains four identical resistors with resistance $ R=2.0\Omega $. Two identical inductors with inductance$ L= 2.0 mH$ and an ideal battery with emf $ E=9V$ The current ‘$ i $just after the switch $ \text{'}s\text{'}$ is closed will be:

L

• A. $9A$
• B. $3A$
• C. $2·25A$
• D. $3·37A$

Answer 1

The correct option is C

$2·25A$

Step 1: Given data

Resistance of identical resistor =$2\Omega$

Inductance in two identical inductor= $2·0mH$

EMF of ideal battery=$9V$

Step 2: Formula used

To determine the current flowing in the circuit, use the Ohm's law $V=iR$ where $V=voltage,i=currentandR=resis\tan ce$.

Step 3: Compute the current $i$ when the switch will be closed

The inductor will behave as an infinite resistance when switch S is closed. As a result, the circuit will resemble

Both the resistances are in serial. So, the equivalent resistance would be,

$\begin{array}{rcl}{R}_{eq}& =& R_1+R_2\\ & =& 2+2\\ & =& 4\Omega \end{array}$

So, the current after switch close,

$$\begin{gathered} i=\frac{9}{4} \\ i=2·25A \end{gathered}$$

So, the current ‘$i$just after the switch $\text{'}s\text{'}$ is closed will be $2·25A$.

Hence, option C is the correct answer.