Question

In the steady-state, find the charge on the capacitor shown in the figure

A) $4\mu C$
B) $5\mu C$
C) $6\mu C$
D) None

Answer 1

We can find the ${R_{equivalent}}$ as the two resistances are connected in series which helps us to find the current flowing through the circuit. This current will help us to find the voltage across the $10\Omega$resistance. After finding the voltage, we can use the formula $Q = CV$.

Formula used:
${R_{series}} = {R_1} + {R_2}$ (when the resistance are in series)
$V = iR$( where $V$ is the voltage across the circuit, $i$ be the current in the circuit and $R$ be the resistance)
$Q = CV$(where $Q$ is the charge of the capacitor, $V$ be the voltage across the capacitor and $C$ be the capacitance of the capacitor)

Complete step by step answer:
We have given a circuit in a steady-state condition. We know that in steady-state conditions no current flows through the capacitor.
In the figure, there are two resistances (${R_1} = 10\Omega$ and ${R_2} = 20\Omega$ ) connected in series. So, the equivalent resistance (${R_{equivalent}}$)will be-
${R_{equivalent}} = 10 + 20 = 30\Omega$
In the given figure, the potential difference ($V = 2V$ )is connected in the circuit. So, the current($i$ ) flow through the circuit is given as-
$i = \dfrac{2}{{30}}{\text{ }}A$
(Where $i = \dfrac{V}{{{R_{equivalent}}}}$)
Now, the voltage drop across the $10\Omega$ resistance is given as,
${V^{'}} = iR$
$\Rightarrow {V^{'}} = \dfrac{2}{{30}} \times 10$
$\Rightarrow {V^{'}} = \dfrac{2}{3}{\text{ }}V$

Which is equal to the voltage across the capacitor.
According to the figure, the charge in the capacitor of capacitance $C = 6\mu C$ will be given as,
$Q = C{V^{'}}$
$Q = 6 \times {10^{ - 6}} \times \dfrac{2}{3}$
$\Rightarrow Q = 2 \times {10^{ - 6}} \times 2$
$\Rightarrow Q = 4 \times {10^{ - 6}}$
$\Rightarrow Q = 4\mu C$

($1\mu C = {10^{ - 6}}C$ )

Hence option (A) is correct.

Note:
In steady-state conditions no current flows through the capacitor. The voltage across the $10\Omega$ resistance is equal to the voltage across the $C = 6\mu C$ capacitance.