Question

Seven capacitors each of capacitance $2\,\mu F$ are to be connected to obtain equivalent capacitance of $\dfrac{10{11}\,\mu F$. Which of the following combination is possible?}

• A. 3 in parallel and 4 in series
• B. 2 in parallel and 5 in series
• C. 5 in parallel and 2 in series
• D. 4 in parallel and 3 in series

Hint:

Always reduce the circuit step by step using series and parallel rules to match the required equivalent capacitance.

Answer 1

The Correct Option is C

Step 1: Capacitance of 5 capacitors in parallel.
Capacitance in parallel adds directly.
\[ C_p = 5 \times 2 = 10\,\mu F \]
Step 2: Capacitance of 2 capacitors in series.
For two identical capacitors in series:
\[ C_s = \frac{2}{2} = 1\,\mu F \]
Step 3: Equivalent capacitance of series combination of $10\,\mu F$ and $1\,\mu F$.
\[ \frac{1}{C_{eq}} = \frac{1}{10} + \frac{1}{1} = \frac{11}{10} \] \[ C_{eq} = \frac{10}{11}\,\mu F \]
Step 4: Conclusion.
Required equivalent capacitance is obtained when 5 capacitors are in parallel and 2 are in series.