The current flowing through the 1 Ω resistor is n/10 A. The value of n is _______.

Solution

Physics sir · 4 steps

Fix the potentials of B and D using the 5 V cell, then one KCL at C and one at A finish the job.

1. Set potentials. Take V_D = 0. The long plate of the 5 V cell faces B, so V_B = 5 V. The long plate of the 10 V cell faces A, so going from C to A through that branch, you lose I₁ × 1 across the resistor and gain 10 V across the cell: V_A = V_C - I₁ + 10.
2. Two KCL equations. At C, the currents coming in from B and D through the 2 Ω resistors leave as I₁. At A, I₁ comes in and leaves through the two 4 Ω resistors.

   I₁ = (5 - V_C)/2 + (0 - V_C)/2 = (V_A - 5)/4 + (V_A - 0)/4

3. Solve. Equating the two KCL forms gives 10 - 4V_C = 2V_A - 5, so V_A = 7.5 - 2V_C. The C to A branch then gives I₁ = V_C - V_A + 10 = 3V_C + 2.5. KCL at C gives I₁ = 2.5 - V_C. Setting these equal gives V_C = 0, and so V_A = 7.5 V.
4. So. I₁ = 2.5 - 0 = 2.5 A = 25/10 A. Check at A: (7.5 - 5)/4 + 7.5/4 = 0.625 + 1.875 = 2.5 A. So n = 25.

The common slip: 2.5. Reporting the current in amperes gives 2.5, but the question asks for n, where I₁ = n/10 A, so n = 25.

The idea to keep. When a cell connects two nodes with no resistor in its branch, it fixes their potential difference directly. Ground one node and use node potentials. You then need only a few KCL equations.

Answer from the NTA final answer key. Solution by PrepWiser.