Question

Figure shows a part of a circuit. If all the capacitors have a capacitance of 2 µF, then the charge on:

• A. C3 is zero
• B. C3 is 12 µC
• C. C₁ is 6 µC
• D. C₂ is 6 µC

Answer 1

Answer: (C), (D)

C₁ is 6 µC, C₂ is 6 µC

Answer 2

Here's how to solve this problem, considering a common interpretation in circuit analysis:

1. Identify the Potentials:

• Leftmost node: 8 V
• Top node: 2 V
• Middle node (connected to C3): 2 V
• Rightmost node: 3 V

2. Initial Assumption and Issue:

• If we assume the junction of C1 and C2 is directly connected to the middle 2V node, then its potential (V_J) is 2V.
• Calculate charges:
  • Q1 = C * (8 - 2) = 2 µF * 6 V = 12 µC
  • Q2 = C * (2 - 2) = 2 µF * 0 V = 0 µC
  • Q3 = C * (2 - 3) = 2 µF * (-1 V) = -2 µC
• None of the options match these values.

3. Alternative Interpretation (Isolated Junction):

• Assume the "2V" label in the middle is only for C3, and the junction (V_J) is an isolated node.
• In this case, the net charge on the isolated plates connected to V_J must be zero.

4. Charge Conservation:

• Charge on the right plate of C1: -Q1 = -C(8 - V_J)
• Charge on the bottom plate of C2: +Q2 = +C(V_J - 2)
• Sum of charges = 0: -C(8 - V_J) + C(V_J - 2) = 0
• Solve for V_J:
  • -(8 - V_J) + (V_J - 2) = 0
  • -8 + V_J + V_J - 2 = 0
  • 2V_J = 10 => V_J = 5 V

5. Recalculate Charges with V_J = 5 V:

• Charge on C1 (Q1): Q1 = C * (8 V - V_J) = 2 µF * (8 V - 5 V) = 2 µF * 3 V = 6 µC
• Charge on C2 (Q2): Q2 = C * (V_J - 2 V) = 2 µF * (5 V - 2 V) = 2 µF * 3 V = 6 µC
• Charge on C3 (Q3): Q3 = C * (2 V - 3 V) = 2 µF * (-1 V) = -2 µC

6. Match with Options:

• With V_J = 5 V:
  • C1 is 6 µC: True
  • C2 is 6 µC: True

Therefore, the intended solution likely assumes the junction is isolated, leading to multiple correct options.