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Question

A capacitor of capacitance C1=1.0μF can with stand a maximum voltage V1=6.0kV. Anothercapacitor of capacitance C2=2.0μF can withstand a maximum voltage V2=4.0kV. If the capacitors are connected in series, the combination can withstand a maximum voltage of


A

10 kV

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B

9 kV

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C

8 kV

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D

6 kV

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Solution

The correct option is B

9 kV


The maximum charge the first capacitor can hold is
Q1=C1V1=1×106×6000=6×103C
The maximum charge the second capacitor can hold is
Q2=C2V2=2×106×4000=8×103C
We know that in a series combination, the charge on each capacitor is the same. Now the first capacitor cannot hold a charge of 8×103C; it can hold a maximum charge of 6×103C. Therefore, the charge on the second capacitor must also be 6×103C. Hence, the voltage across the second capacitor is
V2=6×103C2×106F=3000 volts=3 kilovolts
Thus, the maximum voltage the system can with-stand= V1+V2=6 kilovolts+3 kilovolts=9 kilovolts. Hence the correct choice is (b).


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