Question

If the equivalent capacitance between A and B is $\text{1}\text{F,}$ then the value of C will be

• A. $\text{2}\text{F}$
• B. $\text{4}\text{F}$
• C. $\text{3}\text{F}$
• D. $\text{6}\text{F}$

Answer 1

Answer: (A)

$\text{2}\text{F}$

Answer 2

Capacitors ${{C}_{3}}$ and ${{C}_{4}}$ are in parallel, therefore their resultant capacitance, $C={{C}_{3}}+{{C}_{4}}=1+1=2\mu F$ Now, capacitors ${{C}_{2}}$ and $C$ are in series, therefore their resultant capacitance, $\frac{1}{C\,\,}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}$ $C\,\,=1\mu F$ Capacitors ${{C}_{6}}$ and ${{C}_{8}}$ are in series, therefore their resultant capacitance, $\frac{1}{C\,\,\,\,}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}=1$ $C\,\,\,\,=1\mu F$ Now, $C\,\,\,\,\,\,$ and ${{C}_{5}}$ are in parallel, therefore their resultant capacitance, $C\,\,\,\,\,\,=1+1=2\mu F$ Now, $C\,\,\,\,\,\,$ and ${{C}_{5}}$ are in series. Therefore, their resultant capacitance, $\frac{1}{{{(C)}^{5}}}=\frac{1}{2}+\frac{1}{2}$ Now, $C\,\,$ and ${{(C)}^{5}}$ are in parallel. Therefore, their resultant capacitance, ${{(C)}^{6}}=1+1=2\mu F$ Now, ${{C}_{1}}$ and ${{(C)}^{6}}$ are in series and their resultant capacitance is given $1\mu F$ . $\therefore$ $\frac{1}{1}=\frac{1}{2}+\frac{1}{C}$ $\therefore$ $\frac{1}{C}=\frac{1}{1}-\frac{1}{2}=\frac{1}{2}$ $C=2\mu F$