The electric current in a charging R-C circuit is given by i= $i_{0} e^{- 1 / R C}$ where $i_{0}$ , R and C are constant parameters of the circuit and t is time. Find the rate of change of current at (a) $t = 0$ , (b) t = RC, (c) t = 10 RC.

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Solution

Given that, $i = i_{0 e}^{- v R C}$

$∴$ Rate of change of current

= $\frac{d i}{d t} = i_{0} (\frac{- 1}{RC}) e^{- 1 / R C}$

= $\frac{- i_{0}}{RC} \times e^{- t / R C}$

When

(a) t=0, $\frac{d i}{d t} = \frac{- i_{0}}{RC} \times e^{0} = \frac{- i_{0}}{RC}$

(b) t= RC, $\frac{d i}{d t} = \frac{- i_{0}}{RC} \times e^{- 1} = \frac{- i_{0}}{RCe}$

(c) t= 10 RC, $\frac{d i}{d t} = \frac{- i 0}{R C} \times e^{10}$

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