A 158 mF capacitor with an initial stored energy of 0198 J i
Solution
The initial charge is calculated as follows:
U = Q2/2C
Q = sqrt[2UC] = sqrt[2*0.198*1.58x10-3] = 2.50e-2 C = 25x10-3 C
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Current is calculated as follows:
I = Q/RC = 25x10-3 C / 1.31x106* 1.58x10-3 = 1.2085e-5
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The potential difference:
VC = [Q/C]*e-t/RC
= [25x10-3 /1.58x10-3 ]*e-799/1.31x106* 1.58x10-3
= 10.75 V
= 10.8 V
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The potential difference:
VR = [Q/C]*e-t/RC
= [25x10-3 /1.58x10-3 ]*e-799/1.31x106* 1.58x10-3
= 10.75 V
= 10.8 V
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The rate of thermal energy:
P = [Q2/RC2] * [e-2t/RC]
