Question

In the circuit shown, the charge on $5\mathrm{\mu F}$ the capacitor is:

• A. $5.45\mathrm{\mu C}$
• B. $18.00\mathrm{\mu C}$
• C. $10.90\mathrm{\mu C}$
• D. $16.36\mathrm{\mu C}$

Answer 1

The correct option is D

$16.36\mathrm{\mu C}$

Step 1: Given data

From the given diagram,

$\mathrm{C}=5\mathrm{\mu F}$

${\mathrm{C}}_1=2\mathrm{\mu F}$

${\mathrm{C}}_2=4\mathrm{\mu F}$

Step 2: Calculate the voltage

Knows $\mathrm{Q}=\mathrm{CV}$, where $\mathrm{Q}=$Charge, $C=$Capacitor, and $V=$Voltage.

Therefore, the total charge:

$$\begin{gathered} \Rightarrow \mathrm{Q}={\mathrm{q}}_1+{\mathrm{q}}_2 \\ \Rightarrow \mathrm{C}\times \mathrm{V}={\mathrm{c}}_1\times {\mathrm{v}}_1+{\mathrm{c}}_2\times {\mathrm{v}}_2 \\ \Rightarrow 5\times {\mathrm{V}}_0=2\times \left(6-{\mathrm{V}}_0\right)+4\times \left(6-{\mathrm{V}}_0\right) \\ \Rightarrow 5{\mathrm{V}}_0=12-2{\mathrm{V}}_0+24-4{\mathrm{V}}_0 \end{gathered}$$

Simplify further,

$$\begin{gathered} \Rightarrow 5{\mathrm{V}}_0=36-6{\mathrm{V}}_0 \\ \Rightarrow 5{\mathrm{V}}_0+6{\mathrm{V}}_0=36 \\ \Rightarrow 11{\mathrm{V}}_0=36 \\ \Rightarrow {\mathrm{V}}_0=\frac{36}{11} \end{gathered}$$

Step 3: Calculate the charge on $5\mathrm{\mu F}$capacitor

Now, calculate the charge on $5\mathrm{\mu F}$capacitor

$$\begin{gathered} \mathrm{Q}={\mathrm{CV}}_0 \\ \mathrm{Q}=5{\mathrm{V}}_0 \\ \mathrm{Q}=5\times \frac{36}{11} \\ \mathrm{Q}=\frac{180}{11} \\ \mathrm{Q}=16.36\mathrm{\mu C} \end{gathered}$$

The charge on $5\mathrm{\mu F}$ capacitor is $\mathrm{Q}=16.36\mathrm{\mu C}$

Hence, the option $\left(\mathrm{D}\right)$ is correct.