Question

three identical capacitor C1 C2 and C3 of capacitance 6uf each are connected to a 12V battery (1) charge on each capacitor (2) equivalent capacitance of the network (3) energy stored in the network of capacitor​

Answer 1

Solution:-

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a) Charge in Each Capacitor :

Using KVL(Kirchhoff Voltage Rule) in the loop FCDE:

q3/C3 - 12V = 0

q3 = 12(C3)

q3 = 72 micro C.

Now,

q1 = C1.V and q2 = C2.V

Divide, q1 and q2:

q1/q2 = C1/C2 = 6/6 = 1

=> q1 = q2 = C1.V = C2.V ..................(i)

Now, since the two capacitors C1 and C2 are equal, thus, voltage will be divided equally across C1 and C2

Thus, Voltage across C1 and C2 = 12/2 = 6V

Now,

Using V = 6V in (i),

=> q1 = q2 = C1.V = 6×(6) = 36 micro C.

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b) Equivalent Capacitance of the Network :

It is clear, that Capacitors C1 and C2 are in Series Combination and their equivalent is in Parallel combination with C3.

So,

Solving the series of C1 and C2:

Ceq = (C1.C2)/(C1+C2) = 6×6/(6+6) = 36/12 = 3micro F.

Now, this Ceq is in parallel combination with C3:

Thus,

Ceq' = Ceq + C3 = 6 + 3 = 9 micro F.

Thus, the equivalent capacitance of network is 9 micro Farad.

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c) Energy stored in the capacitors :

Energy stored in C1 = Energy stored in C2 = 1/2(C1.V^2) = 1/2(6×6^2) = 3 × 36 = 108 micro Joule.

and,

Energy stored in C3 = 1/2(C3.V^2) = 1/2(6×12^2) = 3×144 = 432 micro Joules.

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And, that's how you do it!

:)

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