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A capacitor of capacitance 2 μF cannot withstand a voltage more than 5000 V and another capacitor of capacitance 4 μF cannot withstand more than 4000 V. If these capacitors are connected in series, maximum voltage which the combination can withstand is :

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6000 V

5000 V

4000 V

7500 V

answer is D.

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Detailed Solution

Maximum charge on 2 μF = 10000 μC
Maximum charge on 4 μF = 16000 μC
In series connection max. charge = 10000 μC
So maximum voltage of series combination

$= 5000 + \frac{10000 μC}{4 μF} = (5000 + 2500) V = 7500 volt$

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