Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the amount of charge that passes through the battery between t = 0 and t = (L/R) is:

Question

Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the amount of charge that passes through the battery between t = 0 and t = (L/R) is: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/66c7552f459a146cbbb9f4fec5096352-.png" /> (A) (EL/7.3 R2) (B) (EL/2.7 R2) (C) (7.3 EL/ R2) (D) (2.7 EL/ R2)

Tardigrade Admin · Accepted Answer

q = ∫ I dt q = ∫0L/R (E/R) [ 1 - e(-Rt/L)]dt q = (EL/R2) (1/e) q = (EL/2.7 R2)

Q. Consider the $L R$ circuit shown in the figure. If the switch $S$ is closed at $t = 0$ then the amount of charge that passes through the battery between $t = 0$ and $t = \frac{L}{R}$ is:

$\frac{E L}{7.3 R ^{2}}$

$\frac{E L}{2.7 R ^{2}}$

$\frac{7.3 E L}{R ^{2}}$

$\frac{2.7 E L}{R ^{2}}$

$q = \int I d t$
$q = \int_{0}^{L / R} \frac{E}{R} [1 - e^{\frac{- Rt}{L}}] d t$
$q = \frac{E L}{R ^{2}} \frac{1}{e}$
$q = \frac{E L}{2.7 R ^{2}}$

Q. Consider the LR circuit shown in the figure. If the switch S is closed at t=0 then the amount of charge that passes through the battery between t=0 and t=RL​ is:

7.3R2EL​

2.7R2EL​

R27.3EL​

R22.7EL​