Question:
The circuit shown in the figure consists of a charged capacitor of capacity
$3 \mu \mathrm{F}$ and a charge of $30 \mu \mathrm{C}$. At time $\mathrm{t}=0$, when the key is closed, the
value of current flowing through the $5 \mathrm{M} \Omega$ resistor is ' $\mathrm{x}^{\prime} \mu-\mathrm{A}$. The value
of ' $x$ to the nearest integer is
Solution:
(2)
$\mathrm{i}_{0}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{30 / 3}{5 \times 10^{6}}=2 \times 10^{-6}$
$\therefore$ Ans. $=2.00$
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