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.
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
I would not have noticed this if Ms. Sue had not said it was Physics. I never took David College.
C=V/Q, V=CQ,
C=V/Q, V=CQ
To draw the circuit, place the two capacitors in a series connection, which means that one plate of the first capacitor is connected to one plate of the second capacitor. The other plate of the first capacitor should be connected to the positive terminal of the 100V DC supply, and the other plate of the second capacitor is connected to the negative terminal of the supply.
Now, let's calculate the values:
(1) To find the charge on either plate of each capacitor, we know that Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. Since the capacitors are in series, the potential difference across them is the same.
For the first capacitor with a capacitance of 4 mF, the charge can be calculated as Q1 = C1 * V, where C1 = 4 mF and V = 100V.
Q1 = 4 mF * 100V = 400 mC
Similarly, for the second capacitor with a capacitance of 6 mF, the charge can be calculated as Q2 = C2 * V, where C2 = 6 mF and V = 100V.
Q2 = 6 mF * 100V = 600 mC
Thus, the charge on either plate of the first capacitor is 400 milliCoulombs (mC), and the charge on either plate of the second capacitor is 600 mC.
(2) The potential difference across each capacitor can be calculated by using the formula V = Q / C. Since the capacitors are in series, the potential difference across them is the same.
For the first capacitor, using Q1 = 400 mC and C1 = 4 mF:
V1 = Q1 / C1 = 400 mC / 4 mF = 100V
For the second capacitor, using Q2 = 600 mC and C2 = 6 mF:
V2 = Q2 / C2 = 600 mC / 6 mF = 100V
Thus, the potential difference across each capacitor is 100V.
(3) The energy of a capacitor can be calculated using the formula U = 1/2 * C * V², where U is the energy, C is the capacitance, and V is the potential difference.
The total energy of the combined capacitors can be calculated as the sum of the energies of each capacitor.
For the first capacitor, using C1 = 4 mF and V1 = 100V:
U1 = 1/2 * C1 * V1² = 1/2 * 4 mF * (100V)² = 200 mJ
For the second capacitor, using C2 = 6 mF and V2 = 100V:
U2 = 1/2 * C2 * V2² = 1/2 * 6 mF * (100V)² = 300 mJ
Thus, the energy of the combined capacitors is 500 milliJoules (mJ).