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JEE MainPhysicsCurrent Electricity

A total current of 15 mA enters node A of a circuit and splits into two parallel branches that recombine at node B . The first branch consists of a resistor R₁ = 2 k connected to node A and a resistor R₂ = 7 k connected to node B , with a node C between them. The second branch consists of a resistor R₃ = 4 k connected to node A and a resistor R₄ = 2 k connected to node B , with a node D between them. A capacitor of c

Options

  1. A270 C
  2. B120 C
  3. C60 C
  4. D30 C

Correct answer

B. 120 C

Step-by-step solution

In steady state, the capacitor acts as an open circuit, so no current flows through the branch containing it. The total current of 15 mA divides between the two main branches. The equivalent resistance of the first branch is R_ b1 = R₁ + R₂ = 2 k + 7 k = 9 k . The equivalent resistance of the second branch is R_ b2 = R₃ + R₄ = 4 k + 2 k = 6 k . Using the current division rule, the current in the first branch is: I₁ = I_ total R_ b2 R_ b1 + R_ b2 = 15 mA 6 9 + 6 = 6 mA The current in the second branch is: I₂ = I_ to

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