Three capacitors each of capacitance 1,mu F a | vedclass

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(D) Step $1$: Calculate the equivalent capacitance of the three capacitors connected in parallel. For capacitors in parallel,$C_p = C_1 + C_2 + C_3$.

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(D) Step $1$: Calculate the equivalent capacitance of the three capacitors connected in parallel. For capacitors in parallel,$C_p = C_1 + C_2 + C_3$. Given $C_1 = C_2 = C_3 = 1\,\mu F$,we have $C_p = 1 + 1 + 1 = 3\,\mu F$.Step $2$: Calculate the total equivalent capacitance of the system by adding the fourth capacitor in series. For two capacitors in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_4}$.Step $3$: Substituting the values,$\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{1} = \frac{1+3}{3} = \frac{4}{3}$.Step $4$: Therefore,$C_{eq} = \frac{3}{4} = 0.75\,\mu F$.

Three capacitors each of capacitance $1\,\mu F$ are connected in parallel. To this combination,a fourth capacitor of capacitance $1\,\mu F$ is connected in series. The resultant capacitance of the system is.......$\mu F$

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