A capacitor network has two capacitors in series: 150nF and 120nF. The potential difference between point a on the outer side of the 150nF capacitor and point b on the outer side of the 120nF capacitor is 120nF.
a) total charge of the network= 2.4uC
b) charge on each capacitor= 2.4uC, and 2.4uC (in series)
c)total energy in the network=43.2 uJ
d) total energy in each capacitor
i set up two equations V1+V2=36
And C1V1^2/2 + C2V2^2/2 = 43.2
I tried solving but i keep getting the wrong answer
I know the answers are 19.2, and 24.0..
just not sure how to get there
To solve this problem, we can use the relationships between charge, potential difference, and capacitance.
Let's assume the charge on the 150nF capacitor is Q1 and the charge on the 120nF capacitor is Q2. Since the capacitors are in series, the charge on each capacitor will be the same:
Q1 = Q2
We are given that the potential difference between point a and point b is 120V:
V1 + V2 = 120V
The total charge in the network is the sum of the charges on the capacitors:
Total charge = Q1 + Q2
The total energy stored in a capacitor can be calculated using the equation:
E = (Q^2) / (2C)
where E is the energy, Q is the charge, and C is the capacitance. The total energy is the sum of the energies stored in each capacitor:
Total energy = E1 + E2
Now we can substitute the given values into these equations and solve for the unknowns.
First, let's find Q1 and Q2. Since Q1 = Q2, we can simplify the equation to:
2Q1 = Total charge
Given that the total charge is 2.4μC, we have:
2Q1 = 2.4μC
Q1 = Q2 = 1.2μC
Now let's solve for V1 and V2 using the potential difference equation:
V1 + V2 = 120V
Since the capacitors are in series, the total potential difference is equal to the sum of the individual potential differences across each capacitor:
V1 = (Q1) / (C1)
V2 = (Q2) / (C2)
Substituting the values we obtained for Q1, Q2, C1, and C2:
V1 = (1.2μC) / (150nF) = 8V
V2 = (1.2μC) / (120nF) = 10V
Now, let's calculate the total energy in the network using the formula for energy in a capacitor:
E1 = (Q1^2) / (2C1)
E2 = (Q2^2) / (2C2)
Total energy = E1 + E2
Substituting the values we obtained:
E1 = (1.2μC)^2 / (2 * 150nF) = 4.8μJ
E2 = (1.2μC)^2 / (2 * 120nF) = 6μJ
Total energy = 4.8μJ + 6μJ = 10.8μJ
Therefore, none of the provided options a, b, c, or d are correct. The correct answer for the total energy in the network is 10.8μJ.