In a capacitance capacitor C, a charge Q is placed. The capacitor is connected i
ID: 1569181 • Letter: I
Question
In a capacitance capacitor C, a charge Q is placed. The capacitor is connected in the circuit shown in the figure, together with an open circuit breaker, a resistor and a capacitor initially discharged with a capacitance of 3C. The switch is then closed and the circuit is balanced. Determine, depending on Q and C,
A) the final potential difference between the plates of each capacitor,
B) the charge of each capacitor, and
C) the final energy stored in each capacitor.
D) Determine the internal energy that appears in the resistor.
Explanation / Answer
V = Q/C (Total voltage)
After closing the switch the capacitor will be in series then -
Ceq = 3C/4
voltage on C = (V)(3C/4C)
= (Q/C) (3C/4C)
= 3Q/4C
voltage on 3C = (V)(C/4C)
= (Q/C) (C/4C)
= Q/4C
Part b)-
Charge of C = C.(voltage on C)
= 3Q/4
Charge of 3C = 3C.(voltage on 3C)
= 3Q/4
Part C)-
Energy stored in C = 1/2CV2
=1/2 C (3Q/4C)2
= 9Q2 / 32C
Energy stored in 3C = 1/2 (3C) V2
=1/2 *3C *(Q/4C)2
= 3Q2 / 32C
Part D)-
Initially I mention the voltage V = Q/C, this is the voltage across the resister then-
Power dissipated in R = V2 /R
= Q2 / RC2
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