A 70 pF capacitor and a 280 pF capacitor are both charged to 3.6 kV. They are then disconnected from the voltage source and are connected together, positive plate to positive plate and negative plate to negative plate.
Find the resulting potential difference across each capacitor.
kV (70 pF capacitor)
kV (280 pF capacitor)
Find the energy lost when the connections are made.J
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Ответ:
0.636 kJ
Explanation:
The charge on any capacity, q = CV, thus,
The initial charge on the 70 pF capacitor is
q = Cv
q = 70*10^-12 * 3.6*10^3
q = 2.52*10^-7 C
The charge on the 280 pF capacitor is q = C*v
q = 280*10^-12 * 3.6*10^3
q = 1.008x10^-6 C
When they are connected as stated, the net total charge remaining will be 1.008*10^-6 - 2.52*10^-7 = 7.56*10^-7 C
Since the capacitors are in parallel, the equivalent capacitance will be 70 + 280 pF = 350 pF
Remember, q = CV, then V = q/C
V = 7.56*10^-7 C / 350*10^12 F
V = 2160 V
b) The energy before is 1/2 C*v²
E = 1/2 * 70*10^-12 * 3600² + 1/2 * 280*10^-12 * 3600²
E = 4.536*10^-4 J + 1.814*10^-3 J
E = 2.268 kJ
The energy After is 1/2 Cv²
E = 1/2 * 70*10^-12 * 2160² + 1/2 * 280*10^-12 * 2160²
E = 3.266*10^-4 J + 1.306*10^-3 J
E = 1.632 kJ
so the loss is 2.268 - 1.632 = 0.636 kJ
Ответ:
Because mechanical energy (the sum of potential and kinetic energy) is conserved, as the kinetic energy increases, the potential energy decreases. The maximum kinetic energy is achieved when the pendulum passes through the lowest point, and the maximum potential energy is achieved at the highest point.