Asked by abdul waris
two capacitors of capacitance 4 micheal farad and 6 micheal farad are connected in series to a 100v DC supply.Draw the circuit.(A)calculate the:(1)charge on either plate of each capacitor(2)pottential difference accross each capacitor(3)the energy of combined capacitor.
Answers
Answered by
Damon
The charge is the same on each of them. Where else would the electrons on the low side of the top one come from but the high side of the low one :)
C = Q/V for each and the voltages add
so
100 volts = V1 + V2
100 = Q1/C1 + Q2/C2 but Q2=Q1 = Q
so
100 = Q/C1 + Q/C2 = Q (C1+C2)/C1C2
solve for Q
go back and get V1 and V2 from Q/Cn
energy of each = (1/2) C V^2 for each
because
power = i V
so energy = integral i V dt
but C dV/dt = i so i dt = C dV
so energy = integral C V dV
which is (1/2) C V^2
C = Q/V for each and the voltages add
so
100 volts = V1 + V2
100 = Q1/C1 + Q2/C2 but Q2=Q1 = Q
so
100 = Q/C1 + Q/C2 = Q (C1+C2)/C1C2
solve for Q
go back and get V1 and V2 from Q/Cn
energy of each = (1/2) C V^2 for each
because
power = i V
so energy = integral i V dt
but C dV/dt = i so i dt = C dV
so energy = integral C V dV
which is (1/2) C V^2
Answered by
Damon
I would not have noticed this if Ms. Sue had not said it was Physics. I never took David College.
Answered by
value
C=V/Q, V=CQ,
Answered by
Chidiogoezeh
C=V/Q, V=CQ
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