Answer:
a) Q_initial = 16 * 10^-3 C
b) V_1 = V_2 = (16/3) * 10^2 V
c) E = 64/15 J
d) dE = 32/15 J of decrease
Explanation:
Given:
- Capacitor 1, C_1 = 20.0 uF
- Capacitor 2, C_2 = 10.0 uF
- Charged with P.d V = 800 V
Find:
a) the original charge of the system,
(b) the final potential difference across each capacitor
(c) the final energy of the system
(d) the decrease in energy when the capacitors are connected.
Solution:
a)
- The initial charge in the circuit is the one carried by the first charged capacitor.
Q_initial = C_1*V
Q_initial = 20*10^-6 * 800
Q_initial = 16 * 10^-3 C
b)
- After charging the other capacitor, we know that the total charge is conserved among two capacitor:
Q_initial = Q_1 + Q_2
- We also know that potential difference across two capacitor is also same.
V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2
- Using the two equations and solve for charge Q_2:
Q_2 = Q_1*C_2/C_1
Q_2 = Q_1*10/20 = 0.5*Q_1
- using conservation of charge:
Q_initial = 1.5*Q_1
Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C
- Hence the Voltage across each capacitor is:
V_2 = V_1 = Q_1 / C_1
= 10.67*10^-3 / 20*10^-6
= (16/3) * 10^2 V
c)
- The energy in the system is:
E = 0.5*C_eq*V^2
Where, C_eq is the equivalent capacitance of paralle circuit.
E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2
E = 64/15 J
d)
- The decrease in energy of the capacitors is:
dE = E_initial - E_final
Where, E_initial is due to charging of the C_1 only:
dE = 0.5*10^-6*20*800^2 - (64/15)
dE = 32/5 - 64/15 = 32/15 J
