I was absent due to the flu so i\'m at a lost on what to do. Ignore 1 continued.

Question

I was absent due to the flu so i'm at a lost on what to do.  Ignore 1 continued.

L continued: Suppose you connect another capacitor c2 in para||el with the original capacitor C. How does this change your answer to part (d) does it now take more time or less time to charge the pair of capacitors to a given potential (say. 90% of maximum) than it did for only one capacitor? Explain and/or justify your answer. (Thought question (not for credit/. What if you connected C and C2 in series instead?) You have a basic "R-C" circuit without an emf, such as the one shown here. Let R - 8.00 k Ohm and C = 250 nF, and the capacitor is initially charged to a potential of 50.0 V. At time t = 0. the switch is closed and the capacitor begins to discharge. You do NOT need to show your work for fill-in the blank or graphing questions. What is the initial charge on the capacitor, Qn? Qn = What is the initial current that flows in the circuit immediately after the switch is closed? I0 = At what time does the charge on the capacitor reach 1.00% of its initial value? Show your work.

Explanation / Answer

Capacitor discharge (charge decay): Q = Qoe-(t/RC)

so at t = 0, Q = Q0 = cv = 125x10^-7C

b) i = V/R = 6.25mA

c)Q=Q0E^(-t/RC)

Q/Q0 = 0.01

t = -RCln(0.01)

t = 921x10^-5s

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