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(A) For the upper branch,the capacitors are in series. Let the charge be $q_1$. The potential drops are $V_1 = q_1/3$ and $V_2 = q_1/2$. The breakdown

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$2.5$

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(A) For the upper branch,the capacitors are in series. Let the charge be $q_1$. The potential drops are $V_1 = q_1/3$ and $V_2 = q_1/2$. The breakdown conditions are $V_1 \le 1 	ext{ kV}$ and $V_2 \le 2 	ext{ kV}$.$q_1/3 \le 1 \Rightarrow q_1 \le 3 	ext{ mC}$.$q_1/2 \le 2 \Rightarrow q_1 \le 4 	ext{ mC}$.To prevent breakdown,$q_1 \le 3 	ext{ mC}$. The total voltage across the upper branch is $V_{upper} = q_1(1/3 + 1/2) = 3(5/6) = 2.5 	ext{ kV}$.For the lower branch,the capacitors are in series. Let the charge be $q_2$. The potential drops are $V_3 = q_2/7$ and $V_4 = q_2/3$. The breakdown conditions are $V_3 \le 1 	ext{ kV}$ and $V_4 \le 2 	ext{ kV}$.$q_2/7 \le 1 \Rightarrow q_2 \le 7 	ext{ mC}$.$q_2/3 \le 2 \Rightarrow q_2 \le 6 	ext{ mC}$.To prevent breakdown,$q_2 \le 6 	ext{ mC}$. The total voltage across the lower branch is $V_{lower} = q_2(1/7 + 1/3) = 6(10/21) = 60/21 \approx 2.86 	ext{ kV}$.The source voltage $E$ must be the minimum of the two branch voltages to ensure no capacitor breaks down. Thus,$E = \min(2.5, 2.86) = 2.5 	ext{ kV}$.

The diagram shows four capacitors with capacitances and breakdown voltages as mentioned. What should be the maximum value of the external emf source such that no capacitor breaks down? (in $kV$)

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