Question

When the switch is closed, then the initial current through $1\Omega$ resistor is (see Figure)

• A. $12A$
• B. $4A$
• C. $\frac{10}{7}A$
• D. $3A$

Answer 1

The correct option is B $4A$

At $t=0$ the capacitor will behave as a short circuit. The equivalent circuit is shown in the figure.
Now the resistors (6,3) in parallel and combine (6,3) and 1 in series so the equivalent resistance $R_{eq}=\frac{6\times 3}{6+3}+1=2+1=3\Omega$
Current, $I=\frac{V}{R_{eq}}=\frac{12}{3}=4A$
As resistor 1 and (6,3) are in series so current through resistor $1\Omega =4A$