Three capacitors having capacitances of 8.4, 8.4, and 4.2 mFare connected in series across a 36 V potential difference. What is the total energy now stored in the capacitors?

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Three capacitors having capacitances of 8.4, 8.4, and 4.2 mF
are connected in series across a 36 V potential difference. What is the total energy now stored in the
capacitors?

Expert Solution
Step 1 Given

Given,

Capacitor,C1=8.4mF=8.4×10-3FCapacitor,C2=8.4mF=8.4×10-3FCapacitor,C3=4.2mF=4.2×10-3FTotal Voltage,V=36V

We have to calculate the equivalent capacitance,

We have the expression for the capacitor in series connection,

1C=1C1+1C2+1C3

so, 

1C=1038.4+1038.4+1034.21C=10318.4+18.4+14.21C=1030.1190+0.1190+0.23801C=1030.476C=10-30.476C=2.1×10-3F

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