In the circuit shown below, the switch $S$ is connected to position $P$ for a long time so that the charge on the capacitor becomes $q _1 \mu C$. Then $S$ is switched to position $Q$. After a long time, the charge on the capacitor is $q _2 \mu C$.
($1$) The magnitude of $q_1$ is
($2$) The magnitude of $q _2$ is
Give the answer of question ($1$) and ($2$)
IIT 2021, Advanced
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$V _{ A }-1 \cdot i _1-1+2-2 i _1= V _{ A }$
$3 i _1=1$
$i _1=\frac{1}{3} A$
$V _{ A }-1 \cdot i _1-1= V _{ B }$
$V _{ A }- V _{ B }=1+ i _1=\frac{4}{3} \text { volt }$
Potential drop across capacitor $\Delta V =\frac{4}{3}$ volt
$\therefore \quad$ Charge on capacitor $q _1= C \Delta V$
$= 1 \times \frac{4}{3} \mu C$
$q _1= 1.33 \mu C$
(image)
$V _{ A }-1 \cdot i _2+2-2 i _2= V _{ A }$
$3 i _2=2$
$i _2=\frac{2}{3} A$
$V _{ A }- i _2 \times 1= V _{ B }$
$V _{ A }- V _{ B }= i _2 \times 1=\frac{2}{3} \text { volt }$
Potential difference across capacitor $\Delta V =\frac{2}{3}$ volt
$\therefore \quad \text { Charge on capacitor } q _2 = C \Delta V$
$=1 \times \frac{2}{3}=0.67 \mu C$
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