tem 7 Part A A 1.05 F capacitor that is initially uncharged is connected in seri
ID: 1772349 • Letter: T
Question
tem 7 Part A A 1.05 F capacitor that is initially uncharged is connected in series with a 5.03 k resistor and an emf source with 84.2 V and negligible internal resistance. The circuit is completed at t = 0. Just after the circuit is completed, what is the rate at which electrical energy is being dissipated in the resistor? Express your answer with the appropriate units. P-1.41 W Submit My Answers Give Up Correct Significant Figures Feedback: Your answer 1.4 W was either rounded differently or used a different number of significant figures than required for this part. Part B At what value of t is the rate at which electrical energy is being dissipated in the resistor equal to the rate at which electrical energy is being stored in the capacitor? Express your answer with the appropriate units. Submit My Answers Give Up Part C At the time calculated in part B, what is the rate at which electrical energy is being dissipated in the resistor? Express your answer with the appropriate units. P Units Submit My Answers Give Up ContinueExplanation / Answer
current into a cap, charging
v = v[1–e^(–t/)]
v is the battery voltage
i is the current after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = = time constant
Vc = 84.2[1–e^(–t/(1.05µ•5.03k))]
Vc = 84.2[1–e^(–t/(0.0053))]
Vr = 90 – Vc = 84.2 – 84.2[1–e^(–t/(0.0053))]
Vr = 84.2e^(–t/(0.0053))
since the currents in the C and the R are the same, the power (and therefore energy per unit time) is proportional to voltage. So we can set the voltages equal and solve for t
84.2[1–e^(–t/(0.0053))] = 84.2e^(–t/(0.0053))
1–e^(–t/(0.0053)) = e^(–t/(0.0053))
2e^(–t/(0.0053)) = 1
e^(–t/(0.0053)) = 1/2
ln both sides
–t/(0.0053) = –0.693
t = 0.00367 seconds (b)
Vr = 84.2e^(–t/(0.0053))
Vr = 84.2e^(– 0.00367/(0.0053))
Vr = 84.2(0.5) = 42.1 volts
P = Vr²/R = 0.3524 watts
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