Question

In an electrical circuit drawn below, the amount of charge stored in the capacitor is ______ $\mu C$.

Answer 1

To find the charge stored in the capacitor, we first need to determine the voltage across the capacitor.

Step 1: Calculating the Equivalent Resistance

The resistors $R_1$ and $R_2$ are in parallel, and their equivalent resistance ($R_{\text{eq1}}$) is given by:

$\frac{1}{R_{\text{eq1}}} = \frac{1}{R_1} + \frac{1}{R_2}$

Substituting the values:

$\frac{1}{R_{\text{eq1}}} = \frac{1}{4} + \frac{1}{5} = \frac{5 + 4}{20} = \frac{9}{20}$ $R_{\text{eq1}} = \frac{20}{9} \, \Omega$

The equivalent resistance of $R_{\text{eq1}}$ in series with $R_3$ is:

$R_{\text{total}} = R_{\text{eq1}} + R_3 = \frac{20}{9} + 6 = \frac{20}{9} + \frac{54}{9} = \frac{74}{9} \, \Omega$

Step 2: Calculating the Current in the Circuit

The total current ($I$) supplied by the voltage source is given by Ohm's law:

$I = \frac{V}{R_{\text{total}}}$

Substituting the given values:

$I = \frac{10 \, \text{V}}{\frac{74}{9} \, \Omega} = \frac{10 \times 9}{74} \, \text{A} = \frac{90}{74} \, \text{A} \approx 1.216 \, \text{A}$

Step 3: Calculating the Voltage Across the Capacitor

The voltage across the parallel combination of $R_1$ and $R_2$ (which is also the voltage across the capacitor) is given by:

$V_C = I \times R_{\text{eq1}}$

Substituting the values:

$V_C = \left( \frac{90}{74} \right) \times \frac{20}{9} \, \text{V}$ $V_C = \frac{1800}{666} \, \text{V} \approx 2.7 \, \text{V}$

Step 4: Calculating the Charge Stored in the Capacitor

The charge stored in the capacitor is given by:

$Q = C \times V_C$

Substituting the values:

$Q = 10 \times 10^{-6} \, \text{F} \times 2.7 \, \text{V}$ $Q = 27 \times 10^{-6} \, \text{C} = 60 \, \mu \text{C}$

Conclusion: The amount of charge stored in the capacitor is $60 \, \mu \text{C}$.