Question

In the circuit initially switch is closed at time $t=0$ what is the amount of charge passing through the battery in are time constant ?

• A. $\frac{EL}{R^{2}e}$
• B. $\frac{GL}{R^{2}e}$
• C. $\frac{EL}{R^{3}e}$
• D. $\frac{EL}{R^{2}f}$

Answer 1

The correct option is A $\frac{EL}{R^{2}e}$

$\begin{array}{c}ThecurrentflowinginLRcircuitatagiventimetis\\ i=i_0\left(1-e^{-t/T}\right)\\ i_0=\frac{E}{R}andT=\frac{L}{R}\\ cheflowingintimet=T\\ q={\int }_0^{T}idt\\ ={\int }_0^r{i}_0\left(1-e^{-t}r\right)dt\\ =\frac{i_{0}T}{e}\\ =\frac{\frac{E}{R}\frac{L}{R}}{e}\\ =\frac{EL}{R^{2}e}\\ Hence,optionAiscorrectanswer.\end{array}$