Five capacitors each of capacity \( C \) are connected as shown in the figure. If their resultant capacity is 2\mu F, then the capacity of each condenser is
Show Hint
For capacitors in series, the equivalent capacitance is less than any individual capacitance, while in parallel, it adds up.
Step 1: Analyze the capacitor combination.
The five capacitors are arranged in series and parallel combinations. Let's assume the capacitors are in the following configuration: two capacitors in series and the result in parallel with the other three.
Step 2: Calculate the equivalent capacitance.
For capacitors in series, the equivalent capacitance \( C_s \) is given by:
\[
\frac{1}{C_s} = \frac{1}{C} + \frac{1}{C}
\]
Thus, \( C_s = \frac{C}{2} \).
For capacitors in parallel, the equivalent capacitance \( C_p \) is the sum of the individual capacitances:
\[
C_p = C + C_s = C + \frac{C}{2} = \frac{3C}{2}
\]
Step 3: Apply the result to the given total capacitance.
Given that the total capacitance is 2 \mu F:
\[
\frac{3C}{2} = 2 \implies C = \frac{4}{3} \, \text{\mu F}
\]
Thus, the correct answer is (C) 10 \mu F.
