1 ) For Fig 30-23 (on page 694), let C1 = 11.5 ?F, C2 = 4.75 ?F, C3 = 3.95 ?F, a

Question

1 ) For Fig 30-23 (on page 694), let C1 = 11.5 ?F, C2 = 4.75 ?F, C3 = 3.95 ?F, and ?V = 101 V. Now suppose that C3 breaks down electrically, becoming equivalent to a conducting path.

(a) What is the change in the electric charge for capacitor C1? (in C)

(b) What is the change in the potential difference for capacitor C1?? (in V)

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2 ) On The same Figure , let C1 = 12.2 ?F, C2 = 4.82 ?F, C3 = 3.53 ?F, and ?V = 125 V. Now suppose that C3 breaks down electrically, becoming equivalent to a conducting path.?

(a) What is the change in the electric charge for capacitor C1? (in C)

(b) What is the change in the potential difference for capacitor C1?? (in V)?

Explanation / Answer

here,

1)

(C1 + C2) and C3 are in series , their equivalent capacitance , Ceq = (C1 + C2) * C3 /( C1 + C2 + C3)

Ceq = ( 11.5 + 4.75 ) * 3.95 /(11.5 + 4.75 + 3.95) F

Ceq = 3.18 F

the charge across C3 , Q3 = Ceq * V

Q3 = 320.9 C

V1 = (V - Q3/C3) = ( 101 - 320.9 /3.95) = 19.8 V

the charge across C1 , Q1 = V1 * C1

Q1 = 19.8 * 11.5

Q1 = 227.1 C

when C3 breaks

Q1' = C1 * V1 = 1161.5 C

the change in electric charge , dQ = Q1' - Q1 = 934.4 C

b)

the change in the potential difference for capacitor C1 , dV = 101 - 19.8 = 81.2 V

2)

(C1 + C2) and C3 are in series , their equivalent capacitance , Ceq = (C1 + C2) * C3 /( C1 + C2 + C3)

Ceq = ( 12.2 + 4.82 ) * 3.53 /(12.2 + 4.82 + 3.53) F

Ceq = 2.92 F

the charge across C3 , Q3 = Ceq * V

Q3 = 368.75 C

V1 = (V - Q3/C3) = ( 125 - 368.75 /3.53) = 20.5 V

the charge across C1 , Q1 = V1 * C1

Q1 = 20.5 * 12.2

Q1 = 250.6 C

when C3 breaks

Q1' = C1 * V1 = 1525 C

the change in electric charge , dQ = Q1' - Q1 = 1156.3 C

b)

the change in the potential difference for capacitor C1 , dV = 125 - 20.5 = 104.5 V

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