Three capacitors 2 mu F,3 mu F,and 5 mu F can | vedclass

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(B) In a series combination,the charge $q$ on each capacitor is the same.The maximum charge each capacitor can hold is given by $q = C 	imes V_{max}$

Vedclass · Accepted Answer

$(31/6) \, V$

Vedclass · Answer

(B) In a series combination,the charge $q$ on each capacitor is the same.The maximum charge each capacitor can hold is given by $q = C 	imes V_{max}$.For the $2 \, \mu F$ capacitor: $q_1 = 2 \, \mu F 	imes 3 \, V = 6 \, \mu C$.For the $3 \, \mu F$ capacitor: $q_2 = 3 \, \mu F 	imes 2 \, V = 6 \, \mu C$.For the $5 \, \mu F$ capacitor: $q_3 = 5 \, \mu F 	imes 1 \, V = 5 \, \mu C$.To prevent any capacitor from breaking down,the charge in the series combination cannot exceed the minimum of these values,which is $q_{max} = 5 \, \mu C$.The total voltage $V$ that the combination can withstand is the sum of the voltages across each capacitor at this charge:$V = V_1 + V_2 + V_3 = \frac{q_{max}}{C_1} + \frac{q_{max}}{C_2} + \frac{q_{max}}{C_3}$$V = \frac{5 \, \mu C}{2 \, \mu F} + \frac{5 \, \mu C}{3 \, \mu F} + \frac{5 \, \mu C}{5 \, \mu F} = 2.5 + 1.666 + 1 = 5.166 \, V = \frac{31}{6} \, V$.

Three capacitors $2 \mu F$,$3 \mu F$,and $5 \mu F$ can withstand voltages up to $3 \, V$,$2 \, V$,and $1 \, V$ respectively. Their series combination can withstand a maximum voltage equal to.....$V$.

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