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

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Solution

$i = i_{o} e^{- t / R C}$
Rate of change of current at time t = $\frac{d i}{d t} = \frac{- i_{o}}{R C} \times e^{\frac{- t}{R C}}$
(a)Rate of change of current at time t=0 is $\frac{- i_{o}}{R C} \times e^{\frac{0}{R C}} = \frac{- i_{o}}{R C}$
(b)Rate of change of current at time t=RC is $\frac{- i_{o}}{R C} \times e^{\frac{- R C}{R C}} = \frac{- i_{o}}{R C} \times e^{- 1}$
(c)Rate of change of current at time t=10RC is $\frac{- i_{o}}{R C} \times e^{\frac{- 10 R C}{R C}} = \frac{- i_{o}}{R C} \times e^{- 10}$

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