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A capacitor 1mF withstands a maximum voltage of 6KV while another capacitor 2mF withstands a maximum voltage of 4KV. If the capacitors are connected in series, the system will withstand a maximum voltage of

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a

9 KV

b

4 KV

c

2 KV

d

6 KV

answer is D.

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Detailed Solution

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$\begin{array}{l} Q_{1 \max} = 6 \times 10^{3} \times 1 \times 10^{- 3} = 6 C \\ Q_{2 max} = 4 \times 10^{3} \times 2 \times 10^{- 3} = 8 C \\ C_{ef} = \frac{1 \times 10^{- 3} \times 2 \times 10^{- 3}}{3 \times 10^{- 3}} = \frac{2}{3} mF \\ Q_{1 max} = C_{ef} \times V; 6 = \frac{2}{3} \times 10^{- 3} \times V; V = 9 KV \end{array}$

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