A 50 , V potential difference is suddenly app | vedclass

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(D) The current in an $LR$ circuit at time $t$ is given by $i = i_0(1 - e^{-Rt/L})$, where $i_0 = V/R$.The rate of increase of current is $\frac{di}{d

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None of the above

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(D) The current in an $LR$ circuit at time $t$ is given by $i = i_0(1 - e^{-Rt/L})$, where $i_0 = V/R$.The rate of increase of current is $\frac{di}{dt} = \frac{d}{dt} [i_0(1 - e^{-Rt/L})] = i_0 \cdot \frac{R}{L} e^{-Rt/L} = \frac{V}{R} \cdot \frac{R}{L} e^{-Rt/L} = \frac{V}{L} e^{-Rt/L}$.Given $V = 50 \, V$, $L = 5 	imes 10^{-3} \, H$, $R = 180 \, \Omega$, and $t = 0.001 \, s$.Calculate the exponent: $\frac{Rt}{L} = \frac{180 	imes 0.001}{5 	imes 10^{-3}} = \frac{0.18}{0.005} = 36$.Now, $\frac{di}{dt} = \frac{50}{5 	imes 10^{-3}} e^{-36} = 10^4 	imes e^{-36} \, A/s$.Since $e^{-36}$ is an extremely small value, the result is approximately $10^4 	imes 1.7 	imes 10^{-16} \approx 1.7 	imes 10^{-12} \, A/s$.None of the given options match this value.

$A$ $50 \, V$ potential difference is suddenly applied to a coil with $L = 5 \times 10^{-3} \, H$ and $R = 180 \, \Omega$. The rate of increase of current after $0.001 \, s$ is ....... $A/s$.

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