Question

Four capacitors are connected in a circuit as shown in the figure. The effective capacitance between \(P\) and \(Q\) is:

• A. \(10\,\mu F\)
• B. \(5\,\mu F\)
• C. \(2\,\mu F\)
• D. \(7.5\,\mu F\)

Hint:

For capacitor networks: \[ C_{\text{parallel}} = C_1+C_2+\cdots \] \[ C_{\text{series}} = \frac{C_1C_2}{C_1+C_2} \] Always simplify branch-wise and then combine the branches.

Answer 1

The Correct Option is B

Step 1: Identify the parallel combination of the two upper capacitors. The two \(2\,\mu F\) capacitors at the top are connected in parallel. Therefore, \[ C_p = 2 + 2 \] \[ C_p = 4\,\mu F \]

Step 2: Find the equivalent capacitance of the upper branch. The \(4\,\mu F\) capacitor is in series with the \(12\,\mu F\) capacitor. Hence, \[ C_{\text{upper}} = \frac{(4)(12)}{4+12} \] \[ C_{\text{upper}} = \frac{48}{16} \] \[ C_{\text{upper}} = 3\,\mu F \]

Step 3: Combine with the lower branch. The lower branch contains a single \[ 2\,\mu F \] capacitor. The upper and lower branches are connected in parallel between \(P\) and \(Q\). Therefore, \[ C_{\text{eq}} = 3+2 \] \[ C_{\text{eq}} = 5\,\mu F \] Therefore, the effective capacitance between \(P\) and \(Q\) is \[ \boxed{5\,\mu F} \]