Two capacitors of capacitances 1mF and 3mF are charged to the same voltages 5V. They are connected in parallel with oppositely charged plates connected together. Thena)Final common voltage will be 5Vb)Final common voltage will be 2.5 Vc)Heat produced in the circuit will be zerod)Heat produced in the circuit will be 37.5mJCorrect answer is option 'B,D'. Can you explain this answer?

Question

Answer

Explanation:
Concept: Capacitance of a capacitor is defined as the ratio of the magnitude of the charge on either plate of the capacitor to the potential difference across the plates.
C = Q/V
Where C is capacitance, Q is charge, and V is potential difference.

When capacitors are connected in parallel, their equivalent capacitance is the sum of individual capacitances.
Ceq = C1 + C2

When capacitors are connected in parallel, their potential difference is the same and equal to the potential difference of the individual capacitors.

Solution:
Given, Capacitance of capacitor 1, C1 = 1mF = 10^-3 F
Capacitance of capacitor 2, C2 = 3mF = 3 x 10^-3 F
Voltage across both capacitors, V = 5V

Step 1:
Calculate the charge on each capacitor.
Q1 = C1 x V = 10^-3 x 5 = 5 x 10^-3 C
Q2 = C2 x V = 3 x 10^-3 x 5 = 15 x 10^-3 C

Step 2:
When the capacitors are connected in parallel, the equivalent capacitance is
Ceq = C1 + C2 = 1 x 10^-3 + 3 x 10^-3 = 4 x 10^-3 F

Step 3:
The charge on the two capacitors is redistributed when they are connected in parallel. The negative plate of capacitor 1 is connected to the positive plate of capacitor 2, and the positive plate of capacitor 1 is connected to the negative plate of capacitor 2. Therefore, the charges on the two capacitors will combine to give a total charge of zero.

Qtotal = Q1 + Q2 = 5 x 10^-3 C - 15 x 10^-3 C = -10 x 10^-3 C

Step 4:
The final potential difference across the capacitors is the same as the initial potential difference, i.e., 5V.

Step 5:
Calculate the final voltage across the capacitors by using the charge and equivalent capacitance.
Ceq = Qtotal/Vfinal
Vfinal = Qtotal/Ceq
Vfinal = -10 x 10^-3 / 4 x 10^-3 = -2.5V

The final common voltage will be 2.5 V. (Option B)

Step 6:
When the capacitors are connected in parallel, the energy stored in the capacitors will be redistributed. The energy stored in the capacitors before they are connected in parallel is
U1 = 1/2 x C1 x V^2 = 1/2 x 10^-3 x 5^2 = 12.5 x 10^-6 J
U2 = 1/2 x C2 x V^2 = 1/2 x 3 x 10^-3 x 5^2 = 37.5 x 10^-6 J
Total energy stored in the capacitors before they are connected in parallel,
Utotal = U1 + U2 = 12.5 x 10

Answer

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