Question

In the arrangement shown, the charge on capacitor \(C_1\) is: center

center

• A. \(6\mu C\)
• B. \(4\mu C\)
• C. \(8\mu C\)
• D. \(2\mu C\)

Hint:

In series combination of capacitors, charge remains same on every capacitor.

Answer 1

The Correct Option is A

Step 1: Note the capacitor values.
From the circuit, $C_1 = 2\,\mu F$, $C_2 = 2\,\mu F$ and $C_3 = 4\,\mu F$, with a $4$ V battery.

Step 2: Combine the parallel pair.
$C_2$ and $C_3$ are in parallel, so we add them. \[ C_{23} = C_2 + C_3 = 2 + 4 = 6\,\mu F \]
Step 3: Combine with C1 in series.
Now $C_1$ and $C_{23}$ are in series. \[ C_{eq} = \frac{C_1 C_{23}}{C_1 + C_{23}} = \frac{2 \times 6}{2 + 6} = \frac{12}{8} = 1.5\,\mu F \]
Step 4: Find the total charge from the battery.
\[ Q = C_{eq} V = 1.5 \times 4 = 6\,\mu C \]
Step 5: Use the series charge rule.
In a series chain the same charge sits on every capacitor. So the charge on $C_1$ equals this total charge.
Step 6: State the charge on C1.
\[ Q_{C_1} = 6\,\mu C \] \[ \boxed{6\,\mu C} \]