Problem 2: The switch in Fig.2 closes at t = 0s and, after a very long time, the capacitor is fully charged. Find expressions for the total energy supplied by the battery as the capacitor is being charged, total energy dissipated by the resistor as the capacitor is being charged, and the en- ergy stored in the capacitor when it is fully charged. E FIG. 2: The scheme for Problem 2 a) In an ideal battery, a charge dq gains potential energy Edq as it moves up the charge escalator' inside the battery. Therefore, the power supplied by the battery at each moment of time is Pbat = = 18 (this expression was derived in paragraph 28.3 of the textbook). The current in an RC circuit with a charging capacitor changes with time as I = Ioet/t, where Io = &/R and T = RC (this expression was provided at the end of paragraph 28.9 of the textbook). To compute the total energy supplied by the battery as the capacitor is charging, take the integral Ebat = Pbatdt. Express the final answer for Ebat in terms of & and C only. b) The power dissipated by the resistor is equal to Pres= I2R (note that it is a function of time in this problem). Compute the total energy dissipated in the resistor by taking the integral Eres = Predt. Express your answer in terms of & and C only. c) When the capacitor is fully charged, the potential difference between its plates is equal to E, and there is no current flowing in the circuit. Express the potential energy stored in the capacitor in terms of E and C. d) Do your results for parts a) to c) show that the energy is conserved? Explain.
Problem 2: The switch in Fig.2 closes at t = 0s and, after a very long time, the capacitor is fully charged. Find expressions for the total energy supplied by the battery as the capacitor is being charged, total energy dissipated by the resistor as the capacitor is being charged, and the en- ergy stored in the capacitor when it is fully charged. E FIG. 2: The scheme for Problem 2 a) In an ideal battery, a charge dq gains potential energy Edq as it moves up the charge escalator' inside the battery. Therefore, the power supplied by the battery at each moment of time is Pbat = = 18 (this expression was derived in paragraph 28.3 of the textbook). The current in an RC circuit with a charging capacitor changes with time as I = Ioet/t, where Io = &/R and T = RC (this expression was provided at the end of paragraph 28.9 of the textbook). To compute the total energy supplied by the battery as the capacitor is charging, take the integral Ebat = Pbatdt. Express the final answer for Ebat in terms of & and C only. b) The power dissipated by the resistor is equal to Pres= I2R (note that it is a function of time in this problem). Compute the total energy dissipated in the resistor by taking the integral Eres = Predt. Express your answer in terms of & and C only. c) When the capacitor is fully charged, the potential difference between its plates is equal to E, and there is no current flowing in the circuit. Express the potential energy stored in the capacitor in terms of E and C. d) Do your results for parts a) to c) show that the energy is conserved? Explain.
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Hello, I need help with part A and Part B and Part C I just don't understand the problem I was wondering if you help me with the problems and can you label which one is which. Thank you so so so much
Transcribed Image Text:Problem 2: The switch in Fig.2 closes at t = 0s and, after a very long
time, the capacitor is fully charged. Find expressions for the total energy
supplied by the battery as the capacitor is being charged, total energy
dissipated by the resistor as the capacitor is being charged, and the en-
ergy stored in the capacitor when it is fully charged.
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FIG. 2: The scheme for Problem 2
a) In an ideal battery, a charge dq gains potential energy Edq as it
moves up the 'charge escalator' inside the battery. Therefore, the power
supplied by the battery at each moment of time is Phat = = 18 (this
expression was derived in paragraph 28.3 of the textbook). The current
in an RC circuit with a charging capacitor changes with time as I = Ie-t/t, where Io = E/R and T = RC
(this expression was provided at the end of paragraph 28.9 of the textbook). To compute the total energy
supplied by the battery as the capacitor is charging, take the integral Ebat = Pbatdt. Express the final
answer for Ebat in terms of & and C only.
b) The power dissipated by the resistor is equal to Pres= I2R (note that it is a function of time in this
problem). Compute the total energy dissipated in the resistor by taking the integral Eres = Presdt.
Express your answer in terms of & and C only.
c) When the capacitor is fully charged, the potential difference between its plates is equal to E, and
there is no current flowing in the circuit. Express the potential energy stored in the capacitor in terms of
E and C.
d) Do your results for parts a) to c) show that the energy is conserved? Explain.
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