Three capacitors of capacitances C_1=2 mu F,C | vedclass

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(C) The equivalent capacitance $C$ for capacitors in series is given by:$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{2} + \

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Least potential difference is across $C_1$. Equivalent capacitance of combination is $\left(\frac{30}{31}\right) \mu F$. The voltage across $C_3$ is $30 \ V$.

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(C) The equivalent capacitance $C$ for capacitors in series is given by:$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} = \frac{15+10+6}{30} = \frac{31}{30} \mu F^{-1}$Therefore,$C = \frac{30}{31} \mu F$.In a series combination,the charge $q$ on each capacitor is the same:$q = C 	imes V = \left(\frac{30}{31} 	imes 10^{-6} \ Fight) 	imes 155 \ V = 150 	imes 10^{-6} \ C = 150 \mu C$.The potential difference across each capacitor is $V_i = \frac{q}{C_i}$:$V_1 = \frac{150 \mu C}{2 \mu F} = 75 \ V$$V_2 = \frac{150 \mu C}{3 \mu F} = 50 \ V$$V_3 = \frac{150 \mu C}{5 \mu F} = 30 \ V$Comparing the voltages,$V_3 = 30 \ V$ is the least potential difference. Thus,option $C$ is correct.

Three capacitors of capacitances $C_1=2 \mu F$,$C_2=3 \mu F$ and $C_3=5 \mu F$ are connected in series. $A$ potential difference of $155 \ V$ is applied across the combination. Choose the correct option.

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