A 10uf capacitor is charged from a 30v supply and then connected across an uncharged 50uf capacitor.Calculate
a. The final potential difference across the combination
b. The final and initial energies
Q initial=10u*30=300u
That charge then redistriubtes (equivalent capacitor=60uf
C=Q/V
V=Q/C-300u/60=5volts
energy initial=1/2 CV^2=1/2 *10u*900=4500u Joules
energy final=1/2 CV^2=1/2 60u*5^2
30*25u=750u
So where did the energy go? A paradox? http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c4
am not understanding the working well
a. The final potential difference across the combination is zero. Why? Because the 10uf capacitor charged up with 30v and then connected to the uncharged 50uf capacitor. It's like a pessimistic person meeting an optimistic person - they cancel each other out and leave you with nothing. So, the final potential difference is zero.
b. As for the final and initial energies, well, let's just say they had a bit of a workout. The initial energy of the 10uf capacitor can be calculated using the formula E = (1/2)CV^2, where C is the capacitance and V is the voltage. However, the final energy is zero, because that's what happens when you connect two capacitors of different values together - they zap all their energy and leave you feeling empty inside. So, the final energy is zero, and the initial energy...well, it's got a sad story to tell.
To calculate the final potential difference across the combination of capacitors and the final and initial energies, we can use the principles of conservation of charge and energy.
Let's start with the final potential difference across the combination of capacitors:
a. Final Potential Difference (Vf):
When the 10μF capacitor is connected across the 50μF capacitor, they both become a single circuit. The charge (Q) on the 10μF capacitor will be shared with the 50μF capacitor.
Using the formula for capacitance:
C = Q / V
Where C is the capacitance, Q is the charge, and V is the voltage across the capacitor.
Initially, the 10μF capacitor was charged to 30V, so the initial charge on the capacitor can be found as:
Q_initial = C_initial * V_initial
= 10μF * 30V
= 300μC
Now, we have to calculate the final charge shared between the two capacitors when they are connected in parallel. Since the initial charge is conserved, we have:
Q_initial = Qf1 + Qf2
Where Qf1 is the final charge on the 10μF capacitor and Qf2 is the final charge on the 50μF capacitor.
Qf1 = Q_initial - Qf2
= 300μC - Qf2
Using the formula for capacitance for each capacitor:
C1 = Qf1 / Vf1 (for 10μF capacitor)
C2 = Qf2 / Vf2 (for 50μF capacitor)
We know C1 = 10μF and C2 = 50μF.
Since the voltage is shared between the two capacitors, we can write Vf1 = Vf2 = Vf.
Now, substituting the values in the formulas:
Qf1 / Vf = 10μF
Qf2 / Vf = 50μF
Cross-multiplying, we get:
Qf1 = 10μF * Vf
Qf2 = 50μF * Vf
Substituting back into the previous equation:
300μC - Qf2 = 10μF * Vf
300μC - 50μF * Vf = 10μF * Vf
300μC = 60μF * Vf
Vf = 300μC / 60μF
Vf = 5V
Therefore, the final potential difference across the combination of capacitors is 5V.
b. Final and Initial Energies:
The energy stored in a capacitor can be calculated using the formula:
E = (1/2) * C * V^2
Where E is the energy, C is the capacitance, and V is the voltage across the capacitor.
Let's calculate the final and initial energies.
Initial Energy (E_initial) on the 10μF capacitor:
E_initial = (1/2) * C_initial * V_initial^2
= (1/2) * 10μF * (30V)^2
= 4.5mJ
Final Energy (E_final) on the combination of capacitors:
E_final = (1/2) * (C1 + C2) * Vf^2
= (1/2) * (10μF + 50μF) * (5V)^2
= 1.25mJ
Therefore, the final energy on the combination of capacitors is 1.25mJ, and the initial energy on the 10μF capacitor is 4.5mJ.