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(D) The circuit consists of a $12 \, F$ capacitor in series with a parallel combination of two branches,which is then in series with a $16 \, F$ capac

Vedclass · Accepted Answer

$\frac{240}{71} \, F$

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(D) The circuit consists of a $12 \, F$ capacitor in series with a parallel combination of two branches,which is then in series with a $16 \, F$ capacitor.First,consider the upper branch containing $8 \, F$ and $4 \, F$ capacitors in series. Their equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{8} + \frac{1}{4} = \frac{1+2}{8} = \frac{3}{8}$,so $C_1 = \frac{8}{3} \, F$.Next,this $C_1$ is in parallel with the lower branch capacitor of $4 \, F$. The equivalent capacitance of this parallel section $C_p$ is $C_p = C_1 + 4 = \frac{8}{3} + 4 = \frac{8+12}{3} = \frac{20}{3} \, F$.Finally,the total equivalent capacitance $C_{AB}$ is the series combination of $12 \, F$,$C_p = \frac{20}{3} \, F$,and $16 \, F$. Thus,$\frac{1}{C_{AB}} = \frac{1}{12} + \frac{1}{20/3} + \frac{1}{16} = \frac{1}{12} + \frac{3}{20} + \frac{1}{16}$.Finding a common denominator $(240)$: $\frac{1}{C_{AB}} = \frac{20 + 36 + 15}{240} = \frac{71}{240}$.Therefore,$C_{AB} = \frac{240}{71} \, F$.

What is the equivalent capacitance between $A$ and $B$ in the given figure (all values are in farad)?

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