ElectrostaticsHard
Question
Three capacitors C1, C2 and C3 are connected as shown in the figure to a battery of V volt. If the capacitor C3 breaks down electrically the change in total charge on the combination of capacitors is
Options
A.(C1 + C2) V [1 - C3 / (C1 + C2 + C3)]
B.(C1 + C2) V [1 - (C1 + C2) / (C1 + C2 + C3)]
C.(C1 + C2) V [1 + C3 / (C1 + C2 + C3)]
D.(C1 + C2) V [1 - C2 / (C1 + C2 + C3)]
Solution
Equivalent capacitance of circuit,
(Since C1 and C2 are in parallel and parallel and wnich is in seres with C3).
∴ Ceq =
Since V is teh voltage of battery charge, q = Ceq V
If the capacitor C3 breaks down, then effective capacitance,
Ceq = C1 + C2
∴ New charge q′ = C′eq V = (C1 + C2)V
hange in total charge = q′- q
= (C1 + C2) V
= (C1 + C2) V
(Since C1 and C2 are in parallel and parallel and wnich is in seres with C3).
∴ Ceq =
Since V is teh voltage of battery charge, q = Ceq V
If the capacitor C3 breaks down, then effective capacitance,
Ceq = C1 + C2
∴ New charge q′ = C′eq V = (C1 + C2)V
hange in total charge = q′- q
= (C1 + C2) V
= (C1 + C2) V
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