A 10uf capacitor is charged by an 80v supply and then connected across an uncharged 20uf capacitor calculate the final p.d across each capacitor and final charge on each
I guess pd means potential difference (voltage) ?
original
C = Q/V
10*10^-6 = Q / 80
Q = 800 *10^-6
final
total C = 30 *10^-6 and total charge = 800*10^-6
V = Q/C = 800*10^-6 / 30*10^-6 = 80/3 = 26.7 volts
Q1 = 26.7 * 10*10^-6
Q2 = 26.7 * 20*10^-6
I am learning
To find the final potential difference (p.d.) across each capacitor and the final charge on each, we can use the principle of conservation of charge.
1. Initial charge on the 10μF capacitor:
The initial charge on a capacitor can be calculated using the formula: Q = C × V, where Q is the charge, C is the capacitance, and V is the potential difference.
Initial charge Q₁ = 10μF × 80V = 800μC.
2. Final charge on the 10μF capacitor and initial charge on the 20μF capacitor:
Since the two capacitors are connected in parallel, the charge is conserved. Therefore, the final charge on the 10μF capacitor will be equal to the initial charge on the 10μF capacitor.
Final charge Q₁' = 800μC.
The initial charge on the 20μF capacitor is zero since it is initially uncharged.
3. Total charge in the circuit:
The total charge in the circuit is the sum of the charges on each capacitor: Q_total = Q₁ + Q₂.
Q_total = 800μC + 0μC = 800μC.
4. Final charge on the 20μF capacitor:
Since the total charge is conserved, the final charge on the 20μF capacitor can be determined: Q₂' = Q_total - Q₁'.
Q₂' = 800μC - 800μC = 0μC.
5. Final p.d. across each capacitor:
The final potential difference across each capacitor can be calculated using the formula: V = Q / C, where V is the potential difference, Q is the charge, and C is the capacitance.
Final p.d. across the 10μF capacitor: V₁' = Q₁' / 10μF = 800μC / 10μF = 80V.
Final p.d. across the 20μF capacitor: V₂' = Q₂' / 20μF = 0μC / 20μF = 0V.
Therefore, the final p.d. across the 10μF capacitor is 80V, the final p.d. across the 20μF capacitor is 0V, and the final charge on the 10μF capacitor is 800μC, while the final charge on the 20μF capacitor is 0μC.
To calculate the final potential difference across each capacitor and the final charge on each capacitor, we can use the concept of conservation of charge and the formula for capacitors in series.
First, let's calculate the total charge stored in the system before connecting the capacitors. The charge stored in a capacitor is given by the formula:
Q = C * V
Where:
Q = charge stored in the capacitor
C = capacitance of the capacitor
V = potential difference across the capacitor
For the 10µF capacitor charged by the 80V supply:
Q1 = C1 * V1
= (10µF) * (80V)
= 800µC
Since the second capacitor is uncharged initially, its charge is 0µC.
Now, let's calculate the potential difference across each capacitor after connecting them in series. In a series circuit, the charges on the capacitors are equal, and the total potential difference is divided among the capacitors based on their capacitance values. The formula for capacitors in series is:
1/Ceq = 1/C1 + 1/C2
Where:
Ceq = equivalent capacitance of the series combination of capacitors
C1 = capacitance of the first capacitor
C2 = capacitance of the second capacitor
Calculating the equivalent capacitance:
1/Ceq = 1/C1 + 1/C2
= 1/10µF + 1/20µF
= (2 + 1)/20µF
= 3/20µF
Ceq = 20µF/3
Now, let's calculate the potential difference across each capacitor.
Using the proportion of capacitance, we find:
V1/Veq = C1/Ceq
= (10µF) / (20µF/3)
= 3/2
V1 = (3/2) * Veq
Since the capacitors are connected in series, the potential difference across each capacitor will add up to the total potential difference of the system, which is 80V.
So we have:
V1 + V2 = 80V
Substituting the value of V1:
(3/2) * Veq + V2 = 80V
Lastly, let's calculate the final charges on each capacitor using the principle of conservation of charge:
Q1 = Q2
C1 * V1 = C2 * V2
Substituting the values:
(10µF) * ((3/2) * Veq) = (20µF/3) * V2
Now, we have two equations:
(3/2) * Veq + V2 = 80V
(10µF) * ((3/2) * Veq) = (20µF/3) * V2
We can solve these equations simultaneously to find the values of Veq (equivalent potential difference), V2 (potential difference across the second capacitor), and then calculate the final charges on each capacitor.
To solve these simultaneous equations, you can use various methods like substitution, elimination, or matrices.