The circuit shown in the figure consists of a battery of emf varepsilon = 10 ,V ; a capacitor of capacitance C = 1.0 mu

Q: The circuit shown in the figure consists of a battery of $emf$ $\varepsilon = 10 \,V$ ; a capacitor of capacitance $C = 1.0$ $ \mu F$ and three resistor of values $R_1 = 2$ $\Omega$ , $R_2 = 2$ $\Omega$ and $R_3 = 1$ $\Omega$ . Initially the capacitor is completely uncharged and the switch $S$ is open. The switch $S$ is closed at $t = 0.$

d At $t=,$ 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, $I_{R_{1}} R_{1}=I_{R_{2}} R_{2} \Rightarrow \frac{I_{R_{1}}}{I_{R_{2}}}=\frac{R_{2}}{R_{1}}$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

The circuit shown in the figure consists of a battery of $emf$ $\varepsilon = 10 \,V$ ; a capacitor of capacitance $C = 1.0$ $ \mu F$ and three resistor of values $R_1 = 2$ $\Omega$ , $R_2 = 2$ $\Omega$ and $R_3 = 1$ $\Omega$ . Initially the capacitor is completely uncharged and the switch $S$ is open. The switch $S$ is closed at $t = 0.$

AThe current through resistor $R_3 $ at the moment the switch closed is zero.
BThe current through resistor $R_3$ a long time after the switch closed is $5A.$
CThe ratio of current through $ R_1$ and $R_2$ is always constant.
D
all of the above

Diffcult

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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