Question

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

Answer 1

Correct Answer: 60

Given Circuit: 

The circuit consists of a 10V battery, three resistors \( R_1 = 4 \, \Omega \), \( R_2 = 5 \, \Omega \), \( R_3 = 6 \, \Omega \), and a capacitor \( C = 10 \, \mu F \).

In steady state: There will be no current in the branch with the capacitor, so there is no voltage drop across \( R_2 = 5 \, \Omega \), and thus \( I_2 = 0 \).

Step 1: The current in the rest of the circuit:

\[ I_1 = I_3 = \frac{10}{4 + 6} = 1 \, \text{A} \]

Step 2: The voltage drop across \( R_3 \) is given by:

\[ V_{R_3} = V_c + V_{R_2}, \quad V_{R_2} = 0 \]

Step 3: Since no current flows through the capacitor, the voltage across \( R_3 \) is equal to the voltage across the capacitor:

\[ I_3 R_3 = V_c \]

Step 4: Substituting the values:

\[ V_c = 1 \times 6 = 6 \, \text{V} \]

Step 5: The charge on the capacitor is:

\[ q_c = C V_c = 10 \times 6 = 60 \, \mu C \]

Final Answer: The charge on the capacitor is \( 60 \, \mu C \).

Answer 2

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 \( 60 \, \mu \text{C} \).