Q. 253.9( 27 Votes )

# Obtain the equiva

Answer :

Capacitance of capacitor C_{1} = 100pF, C2 = 200pF, C3 = 200 pF, C4 = 100 pF.

Supply potential (V) = 300 V

Let the equivalent capacitance of capacitors C_{2} and C_{3}(connected in series) be C'.

⇒ C’ = 100pF

Let the equivalent capacitance of capacitors C_{1} and C’(connected in parallel) be C’’.

C’’ = C’ + C_{1} = 100 + 100 = 200pF

Let the equivalent capacitance of C’’ and C4 (connected in series) be C’’’.

⇒ C’’’ = 200/3pF

Thus, the equivalent capacitance of the circuit = 200/3 pF.

Now, Supply potential (V) = 300 V

Charge on C_{4} is given by,

Q_{4} = C’’’V = = 2 × 10^{-8}C

So, V_{4} = Q_{4}/C_{4} = (210^{-8})/(10010^{-12}) = 200C

Potential difference across C_{1} is given by, V1 = V-V_{4} = 300-200 = 100V

Charge on C_{1} is given by, Q1 = C_{1}V_{1} = 100 × 10^{-12} × 100 = 10^{-8}C

C_{2} and C_{3} having same capacitances have a potential difference of 100 V together.

Since C_{1} and C_{3} are in series, the potential difference across C_{2} and C_{3} is given by,V_{2} = V_{3} = 50V

Therefore charge on C_{2} is given by,Q_{2} = C_{2}V_{2} = 20010^{-12}50 = 10^{-8}C

And charge on C_{3} is given by,Q_{3} = C_{3}V_{3} = 20010^{-12}50 = 10^{-8}C.

The charge and voltage across each capacitor are as follows:

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