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A capacitor of capacitance $C_{1} = 1 μ F$ can with stand maximum voltage $V_{1} = 6 k V$ (kilo-volt) and another capacitor of capacitance $C_{2} = 3 μ F$ can withstand maximum voltage $V_{2} = 4 k V$ . When the two capacitors are connected in series, the combined system can withstand a maximum voltage of

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$4 k V$

$6 k V$

$8 k V$

$10 k V$

answer is C.

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Detailed Solution

As Q = CV,

( $Q_{1})_{m a x} = 1 0^{- 6} \times 6 \times 1 0^{3} = 6 m C$

While ${(Q_{2})}_{m a x} = 3 \times 1 0^{- 6} \times 4 \times 1 0^{3} = 12 m C$

However in series charge is same so maximum charge on $C_{2}$ will also be 6 mC (and not 12 mC) and potential difference across it $V_{2} = 6 m C / 3 μ F = 2 K V$ and as in series $V = V_{1} + V_{2} s o V_{m a x} = 6 K V + 2 K V = 8 K V$

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