Question

In the given figure, the charge stored in \(6\mu F\) capacitor, when points A and B are joined by a connecting wire is _______\(\mu C\).

Answer 1

Correct Answer: 36

Given: A 9 V DC source. A 6 Ω resistor is in the top left vertical branch. A 3 Ω resistor is in the bottom right vertical branch. A 3 µF capacitor is connected to ground at the bottom left. A 6 µF capacitor is in the right vertical branch. Points A and B are connected by a wire, implying \(V_A=V_B\).

Steady State Behavior: In a DC circuit as time approaches infinity (\(t\to\infty\)), capacitors act as open circuits, blocking steady current. Consequently, only resistors carry current.

1) Circuit Simplification and Current Calculation: With capacitors replaced by open circuits, the only conductive path from the 9 V source to ground is through the series combination of the 6 Ω (top left) and 3 Ω (bottom right) resistors. The equivalent resistance is calculated as: \[ R_{\text{eq}} = 6\,\Omega + 3\,\Omega = 9\,\Omega . \] The total current flowing through this series branch is determined by Ohm's law: \[ i = \frac{V_{\text{source}}}{R_{\text{eq}}} = \frac{9}{9} = 1\ \text{A}. \]

2) Node Voltage Determination: The voltage drop across the 6 Ω resistor is: \[ \Delta V_{6\Omega} = i \cdot 6 = 1 \times 6 = 6\ \text{V}. \] Assuming the top rail is at 9 V and ground is at 0 V, the voltage at node A, located between the 6 Ω resistor and the open-circuited 3 µF capacitor, is: \[ V_A = 9\ \text{V} - \Delta V_{6\Omega} = 9 - 6 = 3\ \text{V}. \] Due to the direct wire connection, the voltage at node B is equal to the voltage at node A: \[ V_B = V_A = 3\ \text{V}. \]

3) Voltage Across the 6 µF Capacitor: The 6 µF capacitor is connected between the 9 V top rail and node B (at 3 V). The potential difference across this capacitor is: \[ \Delta V_{6\mu\mathrm{F}} = 9\ \text{V} - 3\ \text{V} = 6\ \text{V}. \]

4) Charge on the 6 µF Capacitor: The charge stored on the capacitor is calculated using the formula \(Q = C\,\Delta V\): \[ Q = C\,\Delta V = (6\,\mu\text{F}) \times (6\ \text{V}) = 36\,\mu\text{C}. \]

Result: The potential difference across the \(6\,\mu\text{F}\) capacitor is \(6\ \text{V}\), and the stored charge is \[ \boxed{Q = 36\,\mu\text{C}}. \]