Questions

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

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detailed solution

Correct option is D

Maximum charge on 2 μF = 10000 μCMaximum charge on 4 μF = 16000 μCIn series connection max. charge = 10000 μCSo maximum voltage of series combination =5000+10000 μC4 μF =5000+2500V =7500 volt

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