Find the current in the 4 Omega resistance sh | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Just after closing the key,the inductor acts as an open circuit (infinite resistance). The circuit simplifies to a series combination of the $10 \

Vedclass · Accepted Answer

$\frac{5}{3} 	ext{ A}, 0 	ext{ A}$

Vedclass · Answer

(B) Just after closing the key,the inductor acts as an open circuit (infinite resistance). The circuit simplifies to a series combination of the $10 	ext{ V}$ battery,the $2 \Omega$ resistor,and the $4 \Omega$ resistor. The current $I$ is given by $I = \frac{V}{R_{eq}} = \frac{10}{2 + 4} = \frac{10}{6} 	ext{ A} = \frac{5}{3} 	ext{ A}$.$(b)$ $A$ long time after closing the key,the inductor reaches a steady state and acts as a short circuit (zero resistance). Since the inductor is in parallel with the $4 \Omega$ resistor,all the current will flow through the inductor,and the potential difference across the $4 \Omega$ resistor becomes zero. Thus,the current in the $4 \Omega$ resistor is $0 	ext{ A}$.

Find the current in the $4 \Omega$ resistance shown in the figure in the following cases:
$(a)$ Just after the closing of the key.
$(b)$ $A$ long time after the closing of the key.

Similar Questions

In the given circuit,the switch is in position $a$. For what value of $R$ will the circuit have a time constant of $10\,\mu s$ in $K\Omega$?

The total heat produced in resistor $R$ in an $RL$ circuit when the current in the inductor decreases from $I_0$ to $0$ is

Find the time constant of the following circuit.

For the given circuit,the current $i$ through the battery when the key is closed and the steady state has been reached is .....$A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find the current in the $4 \Omega$ resistance shown in the figure in the following cases:$(a)$ Just after the closing of the key.$(b)$ $A$ long time after the closing of the key.