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COMEDK2024Evening ShiftPhysicsCurrent ElectricityActual

Figure below shows a network of resistors, cells, and a capacitor at steady state. What is the current through the resistance 4 ?

Options

  1. A1.0 A
  2. B0.2 A
  3. CZero
  4. D0.5 A

Correct answer

B. 0.2 A

Step-by-step solution

At steady state, capacitor acts as open circuit no current through 3 F branch and the 2 top-left resistor. Simplified circuit has two active branches. Let left node be V_A , right node = 0 V (ground). The 6 V cell sets the middle wire at +6 V. Current through middle branch (via 2 V cell and 1 ): I₁ = (V_A + 2) - 6 1 = V_A - 4 Current through bottom branch (via 3 V cell and 4 ): I₂ = (V_A - 3) - 0 4 = V_A - 3 4 KCL at node A: (V_A - 4) + V_A - 3 4 = 0 4V_A - 16 + V_A - 3 = 0 5V_A = 19 V_A = 3.8 V I_ 4 = 3.8 - 3 4 =

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