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

Idea at steady state: For a DC source as \(t\to\infty\), capacitors behave as open circuits (no steady current through them). Thus only the resistors conduct.

1) Replace capacitors by opens and find the current path
Only the series path through the 6 Ω (left top) and 3 Ω (right bottom) resistors remains between the 9 V terminal and ground. Hence \[ R_{\text{eq}} = 6\,\Omega + 3\,\Omega = 9\,\Omega . \] Total current through the series branch: \[ i = \frac{V_{\text{source}}}{R_{\text{eq}}} = \frac{9}{9} = 1\ \text{A}. \]

2) Node voltages
Voltage drop across the 6 Ω resistor: \[ \Delta V_{6\Omega} = i \cdot 6 = 1 \times 6 = 6\ \text{V}. \] Let the top rail be \(9\ \text{V}\) and ground be \(0\ \text{V}\). The node between the 6 Ω and the (open) 3 µF (point \(A\)) is therefore \[ V_A = 9\ \text{V} - \Delta V_{6\Omega} = 9 - 6 = 3\ \text{V}. \] Since \(A\) and \(B\) are directly connected by a wire, \[ V_B = V_A = 3\ \text{V}. \]

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

4) Charge on the 6 µF capacitor
\[ Q = C\,\Delta V = (6\,\mu\text{F}) \times (6\ \text{V}) = 36\,\mu\text{C}. \]

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