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JEE MainPhysicsCurrent Electricity

In a circuit, nodes B and D are connected to an ideal 12 V battery such that B is at a higher potential. Four resistors are connected as follows: R_ AB = 6 , , R_ AD = 3 , , R_ CB = 3 , , and R_ CD = 6 , . A fifth branch connecting nodes A and C contains a 2 , resistor and a battery of unknown EMF E with its positive terminal connected to node C. If the steady current flowing in the branch from node C to node A is 2

Correct answer

8

Step-by-step solution

Let the potential at node D be V_D = 0 V . Then the potential at node B is V_B = 12 V . Applying Kirchhoff's Current Law (KCL) at node A. The current entering node A from C is 2 A : V_A - 12 6 + V_A - 0 3 = 2 Multiplying the equation by 6: V_A - 12 + 2V_A = 12 3V_A = 24 V_A = 8 V . Applying KCL at node C. The current leaving node C towards A is 2 A : V_C - 12 3 + V_C - 0 6 + 2 = 0 Multiplying the equation by 6: 2V_C - 24 + V_C + 12 = 0 3V_C = 12 V_C = 4 V . For the branch from C to A, the potential difference equat

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