A capacitor 1mF withstands a maximum voltage of 6kV, another 2mF withstands a maximum voltage of 4kV. If they are connected in series, the system withstands a maximum voltage of

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2kV

9kV

4kV

6kV

answer is D.

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Detailed Solution

$\begin{array}{l} q_{1} = C_{1} V_{1} = 6 coulmbs \\ q_{2} = C_{2} V_{2} = 8 coulmbs \end{array}\} If connceted in series, the maximum allowable charge is 6 C V = \frac{q_{1}}{C_{s}} = \frac{6 \times 10^{3}}{2 / 3} = 9 kV$

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