edgarlopez56881 edgarlopez56881
  • 20-07-2018
  • Chemistry
contestada

Given that e = 9.0 v , r = 98 ω and c = 23 μf , how much charge is on the capacitor at time t = 4.0 ms

Respuesta :

ralphwx1 ralphwx1
  • 20-07-2018
Let charge across the capacitor be Q, current through the circuit be I.
Voltage difference across the resistor = rI
Voltage difference across the capacitor = Q/c
Loop rule: net voltage change through a loop must be zero, so
9 = rI + Q/c. Since I = dQ/dt,
r dQ/dt + Q/c = 9
Solving, Q = 9c (1 - e^(t/rc)). Plug in the numbers from the problem for the numerical answer.
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