Question

36. Find the charges on the capacitor shown in figure and the potential difference across them
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Answer 1

Answer:

The total capacitance of the circuit is,

C=2μF

Q=CV

So, Q=240μC

The Voltage across 3μFcapacitor will be,

V1  = 240/3V

= 80 V

The voltage across 2μFand 4μFcapacitor will be,

V2 = [ 120 - 240/3]V

     = 40 V

Charge across the 2μFcapacitor will be,

Q1 =(2×40)μC

   = 80μC

Charge across the 4μFcapacitor will be,

Q2 =(4×40)μC

=160μC

Answer 2

Answer:

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Explanation:

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