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(B) In a series combination,the charge $q$ on each capacitor is the same. Since $q = CV$,we have $V = \frac{q}{C}$,which implies $V \propto \frac{1}{C

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$\frac{11}{3} 	ext{ V}$

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(B) In a series combination,the charge $q$ on each capacitor is the same. Since $q = CV$,we have $V = \frac{q}{C}$,which implies $V \propto \frac{1}{C}$.Given capacitances are $C_1 = 1 \mu	ext{F}$,$C_2 = 2 \mu	ext{F}$,and $C_3 = 3 \mu	ext{F}$.The ratio of voltages across them is $V_1 : V_2 : V_3 = \frac{1}{C_1} : \frac{1}{C_2} : \frac{1}{C_3} = \frac{1}{1} : \frac{1}{2} : \frac{1}{3} = 6 : 3 : 2$.Let $V_1 = 6x$,$V_2 = 3x$,and $V_3 = 2x$.The breakdown voltages are $V_{1,max} = 2 	ext{ V}$,$V_{2,max} = 1 	ext{ V}$,and $V_{3,max} = 3 	ext{ V}$.For the capacitors not to break down,the voltage across each must not exceed its breakdown voltage:$1$) $6x \le 2 \Rightarrow x \le \frac{1}{3} 	ext{ V}$$2$) $3x \le 1 \Rightarrow x \le \frac{1}{3} 	ext{ V}$$3$) $2x \le 3 \Rightarrow x \le \frac{3}{2} 	ext{ V}$To satisfy all conditions,we must choose the smallest value,$x = \frac{1}{3} 	ext{ V}$.The total $emf$ $E = V_1 + V_2 + V_3 = 6x + 3x + 2x = 11x$.Substituting $x = \frac{1}{3} 	ext{ V}$,we get $E = 11 	imes \frac{1}{3} = \frac{11}{3} 	ext{ V}$.

The diagram shows three capacitors with their capacitances and their respective breakdown voltages. What should be the maximum value of the external $emf$ $(E)$ of the source such that no capacitor breaks down?

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