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In the circuit given below, the charge in $\mu \mathrm{C},$ on the capacitor having $5 \mu \mathrm{F}$ is
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9
Potential difference across the branch de is
$6 \mathrm{~V}$. Net capacitance of de branch is $2.1 \mu \mathrm{F}$ So, $q=C V$ $\Rightarrow \mathrm{q}=2.1 \times 6 \mu \mathrm{C} \quad \Rightarrow \mathrm{q}=12.6 \mu \mathrm{C}$
Potential across $3 \mu \mathrm{F}$ capacitance is $\mathrm{V}=\frac{12.6}{3}=4.2 \mathrm{volt}$
Potential across 2 and 5 combination in parallel is $6-4.2=1.8 \mathrm{~V}$
So, $q^{\prime}=(1.8)(5)=9 \mu C$
$6 \mathrm{~V}$. Net capacitance of de branch is $2.1 \mu \mathrm{F}$ So, $q=C V$ $\Rightarrow \mathrm{q}=2.1 \times 6 \mu \mathrm{C} \quad \Rightarrow \mathrm{q}=12.6 \mu \mathrm{C}$
Potential across $3 \mu \mathrm{F}$ capacitance is $\mathrm{V}=\frac{12.6}{3}=4.2 \mathrm{volt}$
Potential across 2 and 5 combination in parallel is $6-4.2=1.8 \mathrm{~V}$
So, $q^{\prime}=(1.8)(5)=9 \mu C$
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