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(C) The given circuit can be simplified by identifying short-circuited components. The capacitors $9\, \mu F, 9\, \mu F$ and $7\, \mu F$ are connected

Vedclass · Accepted Answer

$214 \,\mu C, 27\, V$

Vedclass · Answer

(C) The given circuit can be simplified by identifying short-circuited components. The capacitors $9\, \mu F, 9\, \mu F$ and $7\, \mu F$ are connected in parallel with a wire,effectively short-circuiting them. Thus,they can be removed from the circuit.After simplification,the circuit consists of a parallel combination of $6\, \mu F, 4\, \mu F$ and $8\, \mu F$ capacitors,which is in series with a $36\, \mu F$ capacitor.The equivalent capacitance of the parallel part is $C_p = 6 + 4 + 8 = 18\, \mu F$.The total circuit is now a series combination of $C_p = 18\, \mu F$ and $C_s = 36\, \mu F$ connected to a $40\, V$ source.Let $V_1$ be the potential difference across the $18\, \mu F$ combination and $V_2$ be the potential difference across the $36\, \mu F$ capacitor.We have $V_1 + V_2 = 40\, V$ and the ratio of potential differences in series is inversely proportional to capacitance: $\frac{V_1}{V_2} = \frac{36}{18} = 2$,so $V_1 = 2V_2$.Substituting this into the sum: $2V_2 + V_2 = 40 \implies 3V_2 = 40 \implies V_2 = \frac{40}{3}\, V$.Then $V_1 = 40 - \frac{40}{3} = \frac{80}{3}\, V \approx 26.67\, V$.The potential difference across the $8\, \mu F$ capacitor is $V_1 = \frac{80}{3}\, V \approx 26.67\, V$.The charge on the $8\, \mu F$ capacitor is $Q = C 	imes V_1 = 8\, \mu F 	imes \frac{80}{3}\, V = \frac{640}{3}\, \mu C \approx 213.33\, \mu C$.Rounding to the nearest given option,the values are approximately $214\, \mu C$ and $27\, V$.

In the following diagram,the charge and potential difference across $8\, \mu F$ capacitance will be respectively

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