A current $$I0$$ is flowing through on L – R circuit of time constant $$t0$$ . The source of current is switched off at time t $$=$$ $$0$$. Let r be the value of −dIdt at t $$=$$ $$0$$. Assuming this rate to be constant, the current will reduce to zero in a time interval of

Question

A current $$I0$$  is flowing through on L – R circuit of time constant $$t0$$ . The source of current is switched off at time t $$=$$ $$0$$. Let r be the value of −dIdt at t $$=$$ $$0$$. Assuming this rate to be constant, the current will reduce to zero in a time interval of

Infinity Learn · Accepted Answer

$$I=I0$$  e−$$t/to$$  ⇒  −dIdt  $$=$$ $$I0t0$$  e−$$t/to$$.      ∴    $$r=$$ −$$dIdtt=0$$  $$=$$ $$I0t0$$ ∴              The desired time is $$I0r$$ or $$t0$$

Time -Varying current in L-R circuit

Remember concepts with our Masterclasses.

A current $I_{0}$ is flowing through on L – R circuit of time constant $t_{0}$ . The source of current is switched off at time t = 0. Let r be the value of $(- \frac{dI}{dt})$ at t = 0. Assuming this rate to be constant, the current will reduce to zero in a time interval of

$I = I_{0} e^{- t / t_{o}} \Rightarrow (- \frac{dI}{dt}) = \frac{I_{0}}{t_{0}} e^{- t / t_{o}} . ∴ r = {(- \frac{dI}{dt})}_{t = 0} = \frac{I_{0}}{t_{0}}$
$∴ The desired time is \frac{I_{0}}{r} or t_{0}$

Ready to Test Your Skills?

free study material