Four capacitors are connected as shown in figure. If \( C_1 = 2\,\mu\mathrm{F},\ C_2 = 3\,\mu\mathrm{F},\ C_3 = 5\,\mu\mathrm{F} \) and \( C_4 = 10\,\mu\mathrm{F} \), the equivalent capacitance between point A & B will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 5\,\mu\mathrm{F} \)
\( C_1 \), \( C_2 \) and \( C_3 \) are connected between the same two points, so they are in parallel, and their capacitances add. This group is in series with \( C_4 \).
\( \displaystyle C_{123} = C_1 + C_2 + C_3 = 2 + 3 + 5 = 10\,\mu\mathrm{F} \)
\( \displaystyle C_{AB} = \frac{C_{123}\,C_4}{C_{123} + C_4} = \frac{10 \times 10}{10 + 10} = 5\,\mu\mathrm{F} \)