Two identical capacitors are connected in parallel across a potential ...
Introduction:
When two identical capacitors are connected in parallel across a potential difference (v) and fully charged, the positive plate of the first capacitor is connected to the negative plate of the second capacitor, and the negative plate of the first capacitor is connected to the positive plate of the other capacitor. This arrangement creates a series connection between the capacitors.
Explanation:
When the capacitors are fully charged, they store electrical energy in the form of an electric field between their plates. The energy stored in a capacitor is given by the formula:
E = (1/2) * C * V^2
Where E is the energy stored, C is the capacitance, and V is the potential difference.
Step 1: Capacitors in Parallel:
Initially, when the capacitors are connected in parallel, the potential difference (v) is the same across both capacitors. Therefore, the energy stored in each capacitor is the same and can be denoted as E1 = E2.
Step 2: Capacitors in Series:
When the positive plate of the first capacitor is connected to the negative plate of the second capacitor and vice versa, the capacitors are effectively connected in series. In a series connection, the total capacitance is given by:
C_total = (1/C1) + (1/C2)
However, since the two capacitors are identical (C1 = C2 = C), the total capacitance simplifies to:
C_total = (1/C) + (1/C) = 2/C
The potential difference across the series combination remains the same (v), but the charges on the plates redistribute due to the change in connection.
Step 3: Loss of Energy:
When the capacitors are connected in series, the charge on each capacitor becomes the same. Let's assume this charge to be Q. The potential difference across each capacitor can be calculated using the formula:
V_capacitor = Q/C
Since the potential difference remains the same (v), we have:
v = V1 + V2
v = (Q/C) + (Q/C)
v = (2Q/C)
Using the formula for energy stored in a capacitor, the energy in each capacitor can be calculated as:
E1 = (1/2) * C * (V1^2) = (1/2) * C * [(Q/C)^2] = (1/2) * (Q^2/C)
E2 = (1/2) * C * (V2^2) = (1/2) * C * [(Q/C)^2] = (1/2) * (Q^2/C)
The total energy stored in the series combination can be calculated as the sum of the energies:
E_total = E1 + E2 = (1/2) * (Q^2/C) + (1/2) * (Q^2/C) = (Q^2/C)
Comparing the total energy (E_total) with the initial energy (E1 = E2), we can see that there is no loss of energy in this arrangement. The energy is simply redistributed between the two capacitors in the series combination.
Conclusion:
When two identical capacitors are connected in parallel and then rearranged in a series connection, there is no loss of energy. The
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