Two capacitor in series with 1000v of store of energy of 0.6J while th...
Calculating Capacitance of Two Capacitors in Series and Parallel
Introduction
When capacitors are connected in series or parallel, their combined capacitance changes. In this problem, we are given two capacitors that are connected in series and parallel with a voltage source, and we are asked to calculate their capacitance.
Calculating Capacitance in Series
When capacitors are connected in series, their effective capacitance decreases. The formula for calculating the effective capacitance of capacitors in series is:
1/Ceq = 1/C1 + 1/C2 + ...
Where Ceq is the effective capacitance of the capacitors in series, and C1 and C2 are the capacitances of the individual capacitors.
In this problem, we are given that the two capacitors in series have a store of energy of 0.6J when connected to a 1000V source. We can use this information to calculate the effective capacitance of the capacitors in series.
E = 1/2 * Ceq * V^2
0.6J = 1/2 * Ceq * (1000V)^2
Ceq = 1.2 * 10^-6 F
Therefore, the effective capacitance of the capacitors in series is 1.2 microfarads.
Calculating Capacitance in Parallel
When capacitors are connected in parallel, their effective capacitance increases. The formula for calculating the effective capacitance of capacitors in parallel is:
Ceq = C1 + C2 + ...
Where Ceq is the effective capacitance of the capacitors in parallel, and C1 and C2 are the capacitances of the individual capacitors.
In this problem, we are given that the two capacitors in parallel have a store of energy of 2.5J when connected to the same 1000V source. We can use this information to calculate the effective capacitance of the capacitors in parallel.
E = 1/2 * Ceq * V^2
2.5J = 1/2 * (C1 + C2) * (1000V)^2
C1 + C2 = 5 * 10^-6 F
We also know from the previous calculation that the effective capacitance of the capacitors in series is 1.2 microfarads. We can use this information to calculate the individual capacitances of the two capacitors.
1/Ceq = 1/C1 + 1/C2
1.2 * 10^-6 = 1/C1 + 1/C2
We can solve for one capacitor in terms of the other:
C1 = Ceq / (2 - Ceq/C2)
Substituting this into the equation for the sum of the capacitances:
Ceq / (2 - Ceq/C2) + C2 = 5 * 10^-6
Solving for C2:
C2 = 3.3 * 10^-6 F
Substituting this back into the equation for C1:
C1 = 8.7 * 10^-7 F
Therefore, the capacitance of the two capacitors are 3.3 microfarads