Question 33 of 180

In the given circuit.${C}_{1}=2\mu F,{C}_{2}=0.2\mu F,{C}_{3}=2\mu F,{C}_{4}=4\mu F,{C}_{5}=$
$2\mu F,{C}_{6}=2\mu F$. The charge stored on capacitor ${C}_{4}$ is ______ $\mu C$.

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Answer: 4

From the diagram of the circuit, ${C}_{3},{C}_{4},{C}_{5}$ are in series, so the total capacitance is
$\dfrac{1}{C'}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{2} \Rightarrow C'=\dfrac{4}{5}\mu F$
C'&{C}_{2} are parallel, the total capacitance is
$C''=\left(\dfrac{4}{5}+0.2\right)\mu F=1\mu F$
${C}_{1},C'',{C}_{6}$ are in series. So the equivalent capacitance is,
$\dfrac{1}{{C}_{eq}}=\dfrac{1}{2}+1+\dfrac{1}{2} \Rightarrow {C}_{eq}=0.5\mu F$
Using $Q={C}_{eq}V$.

The charge through the battery, $Q$ $=10\times \dfrac{1}{2}=5\mu C$
From the diagram above,
$Q'=\dfrac{5\mu C\times 0.8}{0.8+0.2}=4\mu C$