Question

Three capacitors each of capacity $4\, \mu F$ are to be connected in such a way that the effective capacitance is $6 \,\mu F$ This can be done by

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

Answer 1

Answer: (A)

connecting two in series and one in parallel

Answer 2

In series order, the net capacitance is, $\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}+\ldots \ldots \ldots$ In parallel order, the net capacitance is, $C=C_{1}+C_{2}+C_{3}+\ldots \ldots \ldots$ We have given, $C_{1}=C_{2}=C_{3}=4 \mu$ (a) The network of three capacitors is shown. Here, $C_{1}$ and $C_{2}$ are in series and the combination of two is in parallel with $C_{3}$. $C_{\text {net }} =\frac{C_{1} C_{2}}{C_{1}+C_{2}}+C_{3}$ $=\left(\frac{4 \times 4}{4+4}\right)+4$ $=2+4=6\, \mu F$ (b) The corresponding network is shown. Here, $C_{1}$. and $C_{2}$ are in parallel and this combination is in series with $C_{3}$. So, $C_{\text {net }}=\frac{\left(C_{1}+C_{2}\right) \times C_{3}}{\left(C_{1}+C_{2}\right)+C_{3}}$ $=\frac{(4+4) \times 4}{(4+4)+4}=\frac{32}{12}$ $=\frac{8}{3} \mu F$ (c) The corresponding network is shown. All of three are in series. So,$\frac{1}{C_{\text {net }}} =\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$ $=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}$ $\therefore C =\frac{4}{3} \mu F$ (d) The corresponding network is shown. All of them are in parallel. So $C_{\text {net }} =C_{1}+C_{2}+C_{3}$ $=4+4+4=12\,\mu F$