Question 25

Obtain the equivalent capacitance of the network in below Figure. For a 300 V supply, determine the charge and voltage across each capacitor.

Answer

Capacitance of capacitor C_{1} is 100 pF.

Capacitance of capacitor C_{2} is 200 pF.

Capacitance of capacitor C_{3} is 200 pF.

Capacitance of capacitor C_{4} is 100 pF.

Supply potential, V = 300 V

Capacitors C_{2} and C_{3} are connected in series. Let their equivalent capacitance be C' .

Capacitors C_{1} and C' are in parallel. Let their equivalent capacitance be

C" and C_{4 }are connected in series. Let their equivalent capacitance be C.

Hence, the equivalent capacitance of the circuit is

Potential difference across C" = V"

Potential difference across C_{4} = V_{4}

Charge on C_{4 }is given by

Q_{4}= CV

Hence, potential difference, V_{1}, across C_{1} is 100 V.

Charge on C1 is given by,

Q_{1} = C_{1}V_{1}

= 100 x 10^{-12} x 100

= 10^{-8} C

C_{2} and C_{3} having same capacitances have a potential difference of 100 V together. Since C_{2} and C_{3} are in series,

the potential difference across C_{2} and C_{3} is given by,

V_{2}= V_{3} = 50 V

Therefore, charge on C_{2} is given by,

Q_{2} = C_{2}V_{2}

= 200 x 10^{-12} x 50

= 10^{-8} C

And charge on C_{3} is given by,

Q_{3} = C_{3}V_{3}

= 200 x 10^{-12} x 50

= 10^{-8} C

Hence, the equivalent capacitance of the given circuit is

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Alfiya
2019-10-15 17:47:52

Solve for 600 v

Deepshikha
2019-10-06 07:31:11

Plzzz solve the rearrangment of circuit

Dimple
2019-08-04 10:39:47

Good description

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