Question

In a circuit containing inductance $L$, resistance $R$ and Voltage $E$,the current in the circuit is given by $L\frac{di}{dt}Ri=E$.The value current in $\frac{L}{R}log2$ seconds is?

• A. Equal to half the maximum value
• B. Equal to maximum value
• C. Equal to one-third of maximum value
• D. Zero

Answer 1

Answer: (A)

Equal to half the maximum value

As in example 1 the current is given by $i=\frac{E}{R}(1-e^{-Rt/L})$

As $t\to \infty ,e^{-Rt/L}\to 0$ and the current reaches its theoretical maximum value say $I$.

$I=\frac{E}{R}$

when $i=I/2$ we get,

$\frac{I}{2}=I(1-e^{-Rt/L})$ $\therefore \frac{1}{2}=1-e^{-Rt/L}$

$\therefore e^{-Rt/L}=\frac{1}{2}$ $\therefore e^{-Rt/L}=2$

$\therefore \frac{Rt}{L}=\log 2$ $\therefore t=\frac{L}{R}\log 2$.