Question

(iii) Three capacitors C₁ = 3 µF, C₂ = 6 µF and C₁ = 6 µF have equal charge q = 30 µC each. C₁ and C₂ are connected in series as shown in figure-3.251. If C3 is connected across the series combination by connecting A with C and B with D and if resistance of connecting wires is 10 Ω, calculate initial current in the circuit just after connections are made and also total amount of heat generated after connections are made.

Answer 1

Initial current = 1 A, Heat generated = 75 µJ

Answer 2

The problem asks for two quantities: the initial current in the circuit just after connections are made, and the total amount of heat generated after connections are made.

1. Initial Current in the Circuit

• Capacitors C₁ and C₂ in series:

  • C₁ = 3 µF, C₂ = 6 µF.
  • Each capacitor has a charge q = 30 µC.
  • Voltage across C₁: V₁ = q/C₁ = 30 µC / 3 µF = 10 V.
  • Voltage across C₂: V₂ = q/C₂ = 30 µC / 6 µF = 5 V.
  • When connected in series (assuming standard polarity where the inner plates cancel out charge), the total voltage across the series combination (points A and B) is V_AB = V₁ + V₂ = 10 V + 5 V = 15 V.
  • The effective capacitance of the series combination is C_series = (C₁C₂)/(C₁+C₂) = (3×6)/(3+6) = 18/9 = 2 µF.
  • The charge on the series combination is Q_series = 30 µC. (This is consistent with V_AB = Q_series / C_series = 30 µC / 2 µF = 15 V).

• Capacitor C₃:

  • C₃ = 6 µF, charge q = 30 µC.
  • Voltage across C₃ (points C and D): V_CD = q/C₃ = 30 µC / 6 µF = 5 V.

• Connection and Initial Current:

  • C₃ is connected across the series combination by connecting A with C and B with D. This means the series combination (C₁-C₂) and C₃ are connected in parallel.
  • At the moment of connection, the potential difference across the C₁-C₂ branch is 15 V, and across the C₃ branch is 5 V. Since they are connected in parallel, these potential differences are not equal, which means there is an immediate potential difference across the connecting wires.
  • Let's assume the positive terminals (A and C) are connected together, and the negative terminals (B and D) are connected together.
  • The effective voltage driving the initial current is the difference between the initial voltages of the two parallel branches: V_effective = |V_AB - V_CD| = |15 V - 5 V| = 10 V.
  • This voltage difference drives a current through the resistance of the connecting wires, R = 10 Ω.
  • Initial current, I_initial = V_effective / R = 10 V / 10 Ω = 1 A.

2. Total Amount of Heat Generated

Heat generated is equal to the difference between the initial total energy stored in the capacitors and the final total energy stored after charge redistribution.

• Initial Total Energy (U_initial):

  • Energy in C₁: U₁ = ½ q²/C₁ = ½ (30 µC)² / (3 µF) = ½ (900/3) µJ = 150 µJ.
  • Energy in C₂: U₂ = ½ q²/C₂ = ½ (30 µC)² / (6 µF) = ½ (900/6) µJ = 75 µJ.
  • Energy in C₃: U₃ = ½ q²/C₃ = ½ (30 µC)² / (6 µF) = ½ (900/6) µJ = 75 µJ.
  • Total initial energy: U_initial = U₁ + U₂ + U₃ = 150 µJ + 75 µJ + 75 µJ = 300 µJ.

• Final Total Energy (U_final):

  • After connection, the capacitors will reach a common potential.
  • The series combination (C₁-C₂) has an effective capacitance C_series = 2 µF and an initial charge Q_series = 30 µC.
  • Capacitor C₃ has a capacitance C₃ = 6 µF and an initial charge Q₃ = 30 µC.
  • When connected in parallel (A to C, B to D, assuming same polarity connection), the total charge is conserved: Q_total = Q_series + Q₃ = 30 µC + 30 µC = 60 µC.
  • The total capacitance of the parallel combination is C_total = C_series + C₃ = 2 µF + 6 µF = 8 µF.
  • The final common potential across the parallel combination is V_final = Q_total / C_total = 60 µC / 8 µF = 7.5 V.
  • Final total energy: U_final = ½ C_total V_final² = ½ (8 µF) (7.5 V)² = 4 × 56.25 µJ = 225 µJ.

• Total Heat Generated (H):

  • The heat generated is the energy lost during charge redistribution: H = U_initial - U_final.
  • H = 300 µJ - 225 µJ = 75 µJ.

The resistance of the connecting wires (10 Ω) is where this heat is dissipated.

Final Answer: The initial current in the circuit just after connections are made is 1 A. The total amount of heat generated after connections are made is 75 µJ.

Explanation of the solution:

1. Calculate initial voltages of C1-C2 series combination and C3. V_series = (30µC/3µF) + (30µC/6µF) = 10V + 5V = 15V. V_C3 = 30µC/6µF = 5V.
2. Initial current is driven by the potential difference between the parallel branches through the connecting wire resistance: I = |15V - 5V| / 10Ω = 10V / 10Ω = 1A.
3. Calculate initial total energy: U_initial = ½(30µC)²/3µF + ½(30µC)²/6µF + ½(30µC)²/6µF = 150µJ + 75µJ + 75µJ = 300µJ.
4. Calculate final common potential: Equivalent capacitance of series C1-C2: C_series = (3×6)/(3+6) = 2µF. Total initial charge (assuming same polarity connection): Q_total = 30µC (from C_series) + 30µC (from C3) = 60µC. Total final capacitance: C_total = C_series + C3 = 2µF + 6µF = 8µF. Final common potential: V_final = Q_total / C_total = 60µC / 8µF = 7.5V.
5. Calculate final total energy: U_final = ½ C_total V_final² = ½ (8µF) (7.5V)² = 225µJ.
6. Heat generated: H = U_initial - U_final = 300µJ - 225µJ = 75µJ.