Three capacitors, each of capacitance 120 pF, are each charged to 500 V and then connected in series. Determine (a) the potential difference between the end plates, (b) the charge on each capacitor, and (c) the energy stored in the system. Ans. a.) 1.5kV b.)60nC c.) 45*10^-6 J
I NEED THE SOLUTION ON HOW TO GET THIS PLLLLLSSS.
C = Q/V
If each is charged to 500 volts and they are connected in series, then nothing happens, they each remain at 500 volts because no current flows.
The total voltage is 3 * 500 = 1500 Volts
Q = C V = 120 * 10^-12 * 500 = 60,000*10^-12 = 60^10^-9 farad
energy in EACH = (1/2) Q V = 30*10^-9 * (500) = 15 * 10^-6
multiply by 3 = 45 *10^-6 Joules
a. V = 3 * 0.5kv =
b. Q = C*V = 120*10^-12 * 500 = 60*10^3 * 10^-12 = 60^10^-9 C = 60 nC.
c. Ceq = 120/3 = 40 pF.
Eneroy = Ceq*V/2 = 40*10^-12 * 1500^2/2 = 45*10^-6 J.
Haw did you get the (3)?
Oh, I see you're in need of some electrifying information! Well, let's break it down and zap through the steps to find our answers:
(a) The potential difference between the end plates in a series connection of capacitors is equal to the sum of the individual potential differences. Since each capacitor is charged to 500 V, the total potential difference between the end plates would be 500 V + 500 V + 500 V = 1500 V, which can also be expressed as 1.5 kV.
(b) In a series connection of capacitors, the charge on each capacitor is the same. So, the charge on each capacitor would be the total charge divided by the number of capacitors. Since we don't have the total charge mentioned in the question, we'll need to calculate it. We can use the formula Q = C × V, where Q is charge, C is capacitance, and V is voltage.
For each capacitor, Q = (120 pF) × (500 V) = 60 nC. Therefore, the charge on each capacitor is 60 nC.
(c) The energy stored in a capacitor can be calculated using the formula U = 0.5 × C × V^2, where U is energy, C is capacitance, and V is voltage.
For each capacitor, U = 0.5 × (120 pF) × (500 V)^2 = 45 × 10^(-6) J. Since we have three capacitors, the total energy stored in the system would be 3 × 45 × 10^(-6) J = 135 × 10^(-6) J, which can also be expressed as 0.135 J.
So there you have it! The potential difference between the end plates is 1.5 kV, the charge on each capacitor is 60 nC, and the energy stored in the system is 45 × 10^(-6) J, or 0.135 J if you prefer a more electrifying expression.
To find the potential difference between the end plates when the capacitors are connected in series, we need to use the formula for the total capacitance of capacitors in series.
The formula for the total capacitance (C_total) of capacitors in series is given by:
1/C_total = 1/C1 + 1/C2 + 1/C3 + ...
In this case, we have three capacitors with capacitance C = 120 pF each. Substituting the values into the formula, we get:
1/C_total = 1/120 pF + 1/120 pF + 1/120 pF
Since all the values are the same, we can simplify the equation to:
1/C_total = 3/120 pF
To find C_total, we can take the reciprocal of both sides:
C_total = 120 pF/3
C_total = 40 pF
Now that we have the total capacitance, we can find the potential difference between the end plates using the formula for capacitors in series:
C_total = (Q_total) / (V_total)
Where Q_total is the total charge stored and V_total is the potential difference between the end plates.
Substituting the values into the formula, we get:
40 pF = Q_total / 500 V
To find Q_total, we can rearrange the equation:
Q_total = C_total * V_total
Q_total = 40 pF * 500 V
Q_total = 20,000 pC (pC = picocoulombs)
Converting picocoulombs to nanocoulombs (nC):
Q_total = 20,000 pC * (1/1000) nC/pC
Q_total = 20 nC
Therefore, the charge on each capacitor is 20 nC.
Next, let's find the energy stored in the system. The energy stored in a capacitor can be calculated using the formula:
E = (1/2) * C_total * (V_total)^2
Substituting the values into the formula, we get:
E = (1/2) * 40 pF * (500 V)^2
E = (1/2) * 40 pF * 250,000 V^2
E = 20,000 pJ (pJ = picojoules)
Converting picojoules to joules:
E = 20,000 pJ * (1/10^12) J/pJ
E = 20 * 10^-6 J
Therefore, the energy stored in the system is 20 * 10^-6 J or 20 μJ (microjoules).
In summary:
(a) The potential difference between the end plates is 1.5 kV.
(b) The charge on each capacitor is 60 nC.
(c) The energy stored in the system is 45 * 10^-6 J or 45 μJ.