Question

All capacitors were initially uncharged. Then,

• A. Current from the battery just after closing of switch S is $3.42\text{A}$
• B. Current from the battery just after closing of switch S is $0.96\text{A}$
• C. Current from the battery long time after of closing of the switch S is $3.42\text{A}$
• D. Current from the battery after long time of closing of switch S is $0.96\text{A}$

Answer 1

Answer: (A), (D)

Current from the battery after long time of closing of switch S is $0.96\text{A}$

Just after closing of switch S, capacitor acts as conducting wire



$\frac{1}{R_{eq}}=\frac{1}{15}+\frac{1}{15}+\frac{1}{12}$
$\frac{1}{R_{eq}}=\frac{2}{15}+\frac{1}{12}$
$R_{eq}=\frac{180}{39}=4.61$
As these resistances are in series with $10\Omega$
$R_{net}=4.61+10=14.61$
$\therefore i=\frac{50}{14.61}=3.42A$

After closing of switch S, for a long time, capacitor acts as open circuit

We can observe that all resistances are in series,
$R_{net}=15+15+12+10=52\Omega$

$\therefore i=\frac{50}{52}=0.96\text{A}$