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

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

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

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

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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

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$t_{0}$

${rt}_{0}$

$t_{0} / r$

$(1 - \frac{1}{r}) t_{0}$

answer is A.

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

$∴ The desired time is \frac{I_{0}}{r} or t_{0}$

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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

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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