Question

The circuit shown in the figure consists of a battery emf $\varepsilon=10\, V ;$ a capacitor of capacitance $C=1.0\, \mu F$ three resistors of values $R_{1}=2\, \Omega, R_{2}=2\, \Omega$ and $R_{3}=1$ Initially the capacitor is completely uncharged and switch $S$ is open. The switch $S$ is closed at $t=0 .$ Then

• A. the current through resistor $R_{3}$ at the moment switch closed is zero
• B. the current through resistor $R_{3}$ a long time after switch closed is $5 A$
• C. the ratio of current through $R_{1}$ and $R_{2}$ is alw constant
• D. all of these

Answer 1

Answer: (D)

At $t=\infty$, the equivalent circuit is
$I=\frac{\varepsilon}{R_{3}+\frac{R_{1} R_{2}}{R_{1}+R_{2}}}$
$=\frac{10}{1+\frac{(2)(2)}{2+2}}=\frac{10}{1+1}=5 \,A$

Also, $I_{R_{1}} R_{1}=I_{R_{2}} R_{2} $
$\Rightarrow \frac{I_{R_{1}}}{I_{R_{2}}}=\frac{R_{2}}{R_{1}}=$ constant