In the circuit given below, what is the charge in $\mu \mathrm{C}$, on the capacitor having $5 \mu \mathrm{F} ?$

Question

Aicharya · Accepted Answer

Potential difference across the branch de is $6 \mathrm{~V}$. Net capacitance of de branch is $2.1 \mu \mathrm{F}$ So, $\mathrm{q}=\mathrm{CV}$
$\Rightarrow \mathrm{q}=2.1 \times 6 \mu \mathrm{C}$
$\Rightarrow \mathrm{q}=12.6 \mu \mathrm{C}$
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, $q^{\prime}=(1.8)(5)=9 \mu \mathrm{C}$

In the circuit given below, what is the charge in $\mu \mathrm{C}$, on the capacitor having $5 \mu…

Solution