Question

Four capacitors are connected as shown in the figure. If \(C_1=2\,\mu F\), \(C_2=3\,\mu F\), \(C_3=5\,\mu F\) and \(C_4=10\,\mu F\), then the equivalent capacitance between points \(A\) and \(B\) is:

• A. \(20\,\mu F\)
• B. \(5\,\mu F\)
• C. \(10\,\mu F\)
• D. \(0\,\mu F\)

Hint:

For capacitors: Parallel: \[ C=C_1+C_2+C_3+\cdots \] Series: \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} +\cdots \] Always simplify parallel combinations first whenever they share the same two nodes.

Answer 1

The Correct Option is B

Step 1: Identify the parallel combination. The left and right junctions are common for \(C_1\), \(C_2\), and \(C_3\). Therefore, these three capacitors are connected in parallel. Hence, \[ C_p=C_1+C_2+C_3 \] \[ C_p=2+3+5 \] \[ C_p=10\,\mu F \]

Step 2: Identify the series combination. The equivalent capacitor \(C_p=10\,\mu F\) is in series with \[ C_4=10\,\mu F \] For capacitors in series: \[ C_{\text{eq}} = \frac{C_pC_4}{C_p+C_4} \] \[ C_{\text{eq}} = \frac{(10)(10)}{10+10} \] \[ C_{\text{eq}} = \frac{100}{20} \] \[ C_{\text{eq}} = 5\,\mu F \] Therefore, the equivalent capacitance between \(A\) and \(B\) is \[ \boxed{5\,\mu F} \]