Question

Three capacitors each of capacity 4µF are to be connected in such a way that the effective capacitance of 6µF. This can be done by

• A. connecting all of them in a series
• B. connecting them in parallel
• C. connecting two in series and one in parallel
• D. connecting two in parallel and one in series

Answer 1

The Correct Option is C

To find the configuration that results in an effective capacitance of 6µF using three capacitors, each with a capacitance of 4µF, let us explore the options:

1. Connecting all of them in series: In a series connection, the total or effective capacitance C_s is given by: \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}.
   For three capacitors of 4µF each, \frac{1}{C_s} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4} \Rightarrow C_s = \frac{4}{3} \text{µF}.
   This does not yield the required 6µF.
2. Connecting all of them in parallel: In a parallel connection, the total capacitance C_p is the sum: C_p = C_1 + C_2 + C_3.
   Therefore, C_p = 4 + 4 + 4 = 12 \text{µF}.
   This, too, does not result in 6µF.
3. Connecting two in series and one in parallel: First, connect two capacitors in series: \frac{1}{C_s} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \Rightarrow C_s = 2 \text{µF}.
   Then, connect this series combination in parallel with the third capacitor: C_{\text{eff}} = C_s + C_3 = 2 + 4 = 6 \text{µF}.
   This configuration gives the desired effective capacitance of 6µF.
4. Connecting two in parallel and one in series: Two capacitors in parallel yield: C_p = 4 + 4 = 8 \text{µF}.
   Connecting this parallel combination in series with the third capacitor, we get: \frac{1}{C_{\text{eff}}} = \frac{1}{8} + \frac{1}{4} = \frac{3}{8} \Rightarrow C_{\text{eff}} = \frac{8}{3} \text{µF}.
   This configuration results in a capacitance of approximately 2.67µF, which is not 6µF.

Hence, connecting two capacitors in series and one in parallel is the correct option, yielding the effective capacitance of 6µF.