Four capacitors are connected as shown in the figure. Here \( C_1=2\,\mu\mathrm{F} \), \( C_2=3\,\mu\mathrm{F} \), \( C_3=6\,\mu\mathrm{F} \) and \( C_4=2\,\mu\mathrm{F} \).
What is the equivalent capacitance between A and B?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 3\,\mu\mathrm{F} \)
Capacitors joined one after another on a single path are in series, with \( 1/C=1/C_1+1/C_2+\ldots \). Capacitors connected across the same two points are in parallel, with \( C=C_1+C_2+\ldots \). From the figure, \( C_1 \) is connected directly between A and B. The other path from A to B passes through \( C_2 \), \( C_3 \) and \( C_4 \) one after another.
So \( C_2 \), \( C_3 \) and \( C_4 \) are in series, and their effective capacitance \( C' \) is:
\( \begin{aligned} \frac{1}{C'} &= \frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4} \\ &= \frac{1}{3\,\mu\mathrm{F}}+\frac{1}{6\,\mu\mathrm{F}}+\frac{1}{2\,\mu\mathrm{F}} \\ &= \frac{2+1+3}{6\,\mu\mathrm{F}} \\ &= \frac{1}{1\,\mu\mathrm{F}} \end{aligned} \)
\( C'=1\,\mu\mathrm{F} \)
This series branch is in parallel with \( C_1 \) between A and B:
\( \begin{aligned} C_{AB} &= C_1+C' \\ &= 2\,\mu\mathrm{F}+1\,\mu\mathrm{F} \\ &= 3\,\mu\mathrm{F} \end{aligned} \)