Question 74: Electrostatics & Capacitance

Question

The stored charge in the capacitor in steady state of the following circuit is $\_\_\_\_$ $\mu \mathrm{C}$.

Accepted Answer

When a capacitor is connected to a DC source, it begins to charge. In steady state the capacitor is fully charged, so it acts as an open circuit, i.e., no current flows through the branch containing the capacitor. Current only flows through the closed resistive loops of the circuit.
To find the voltage across the capacitor, we first need the total current from the 12 V battery.
Since no current flows through the capacitor, the $\mathrm{R}_7=10 \Omega$ resistor in that branch also carries no current.
The resistors $\mathrm{R}_8=4 \Omega, \mathrm{R}_4=4 \Omega$ and $\mathrm{R}_6=2 \Omega$ are in series
$$ \mathrm{R}_{4,6,8}=\mathrm{R}_4+\mathrm{R}_6+\mathrm{R}_8=(4+4+2) \Omega=10 \Omega $$
The resistor $\mathrm{R}_5=10 \Omega$ is in parallel with the combination $\mathrm{R}_{4,6,8}$.
$$ \begin{aligned} \frac{1}{\mathrm{R}_{5,(4,6,8)}}= & \frac{1}{\mathrm{R}_5}+\frac{1}{\mathrm{R}_{4,6,8}}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10} \ & \Rightarrow \mathrm{R}_{5,(4,6,8)}=5 \Omega \end{aligned} $$
Now, $\mathrm{R}_2=5 \Omega, \mathrm{R}_{5,(4,6,8)}=5 \Omega$, and $\mathrm{R}_3=2 \Omega$ are in series.
$$ \mathrm{R}_{2,(5,(4,6,8)), 3}=\mathrm{R}_2+\mathrm{R}_{5,(4,6,8)}+\mathrm{R}_3=5 \Omega+5 \Omega+2 \Omega=12 \Omega $$
This $\mathrm{R}_{2,(5,(4,6,8)), 3}=12 \Omega$ is in parallel with $\mathrm{R}_1=12 \Omega$.
$$ \mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}_{2,(5,(4,6,8)), 3} 	imes \mathrm{R}_1}{\mathrm{R}_{2,(5,(4,6,8)), 3}+\mathrm{R}_1}=\frac{12 	imes 12}{12+12}=6 \Omega $$
The total resistance across 12 V battery is $\mathrm{R}_{\mathrm{eq}}=6 \Omega$.
So, using Ohm's law the total current from source is,
$$ I_{	ext {total }}=\frac{V}{R_{\mathrm{eq}}}=\frac{12}{6}=2 \mathrm{~A} $$
This current divides equally between $\mathrm{R}_1$ and the rest of the circuit (since both branches are $12 \Omega$ ).
Current entering the $\mathrm{R}_2$ branch $\mathrm{I}_2=1 \mathrm{~A}$.
This 1 A current reaches the node at $\mathrm{R}_5$ and divides between $\mathrm{R}_5=10 \Omega$ and the $\mathrm{R}_4-\mathrm{R}_8-\mathrm{R}_6$ branch $(10 \Omega)$.
So, current through $\mathrm{R}_8$ is $\mathrm{I}_8=0.5 \mathrm{~A}$.
The potential difference across the capacitor is equal to the potential difference across the branch it is connected to in parallel.
$$ \begin{gathered} \mathrm{V}_{\mathrm{C}}=\mathrm{I}_8 	imes \mathrm{R}_8 \ \mathrm{~V}_{\mathrm{C}}=0.5 	imes 4=2 \mathrm{~V} \end{gathered} $$
The formula for charge stored in a capacitor is :
$$ \mathrm{Q}=\mathrm{CV}_{\mathrm{C}} $$
The capacitance is $\mathrm{C}=100 \mu \mathrm{~F}=100 	imes 10^{-6} \mathrm{~F}=10^{-4} \mathrm{~F}$ and the voltage across it is $\mathrm{V}_{\mathrm{C}}=2 \mathrm{~V}$ Putting the values,
$$ Q=\left(10^{-4}ight) 	imes 2 $$
$$\Rightarrow $$ $$ Q=2 	imes 10^{-4} C=200 	imes 10^{-6} C $$
Therefore, the energy stored in the capacitor at steady state is $200 \mu \mathrm{C}$.

Electrostatics & Capacitance

The stored charge in the capacitor in steady state of the following circuit is \_\_\_\_ \mu \mathrm{C}.

Integer / Numeric Answer Type