The electric current in a charging R-C circuit is given by i=i_0e… - Vidyadip

Question

The electric current in a charging $R-C$ circuit is given by
$\text{i}=\text{i}_0\text{e}^{-\frac{\text{t}}{\text{RC}}}$
where $\text{i}_0, R$ and $C$ are constant parameters of the circuit and $t$ is time.
Find the rate of change of current at

$t = 0.$
$t = RC.$
$t = 10RC.$

✓

Answer

Given that, $\text{i}=\text{i}_0\text{e}^{-\frac{\text{t}}{\text{RC}}}$
$\therefore$ Rate of change of current $=\frac{\text{di}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\text{i}_0\text{e}^{-\frac{\text{i}}{\text{RC}}}=\text{i}_0\frac{\text{d}}{\text{dt}}\text{e}^{-\frac{\text{t}}{\text{RC}}}=\frac{-\text{i}_0}{\text{RC}}\times\text{e}^{-\frac{\text{t}}{\text{RC}}}$

When $\text{t}=0,\frac{\text{di}}{\text{dt}}=\frac{-\text{i}}{\text{RC}}$
when $\text{t}=\text{RC},\frac{\text{di}}{\text{dt}}=\frac{-\text{i}}{\text{RCe}}$
when $\text{t}=10\text{RC},\frac{\text{di}}{\text{dt}}=\frac{-\text{i}}{\text{RCe}^{10}}$

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