Consider the circuit shown in figure. The current through the battery a long time after the switch S is closed is

Question

Consider the circuit shown in figure. The current through the battery a long time after the switch S is closed is (A) (E/R1) (B) (E/R2) (C) (E/R1+R2) (D) (E(R1+R2)/R1 R2)

Tardigrade Admin · Accepted Answer

After a long time, steady current flows, so no emf is induced across inductor. Hence, circuit reduces like shown. So, current flowing, i=(E(R1+R2)/R1 R2) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/67d861a6521c6775f44a4add415b64e1-.png" />

Q. Consider the circuit shown in figure. The current through the battery a long time after the switch $S$ is closed is

$\frac{E}{R _{1}}$

$\frac{E}{R _{2}}$

$\frac{E}{R _{1} + R _{2}}$

$\frac{E ( R _{1} + R _{2} )}{R _{1} R _{2}}$

After a long time, steady current flows, so no emf is induced across inductor.
Hence, circuit reduces like shown. So, current flowing,
$i = \frac{E ( R _{1} + R _{2} )}{R _{1} R _{2}}$

Q. Consider the circuit shown in figure. The current through the battery a long time after the switch S is closed is

R1​E​

R2​E​

R1​+R2​E​

R1​R2​E(R1​+R2​)​