A main circuit branch PQ carries a constant current i₁ = 2 A from P to Q. The branch contains a resistor of 5 and a primary coil of self-inductance 4 H . A secondary coil is magnetically coupled to the primary coil with a mutual inductance of 3 H . If the current i₂ in the secondary coil is increasing at a rate of 2 A/s and the resulting mutually induced emf opposes the current i₁ in the primary coil, what is the pot

Step-by-step solution

Apply Kirchhoff's Voltage Law (KVL) from P to Q. The current i₁ is constant, so the self-induced emf in the primary coil is zero ( L di₁ dt = 0 ). The mutually induced emf in the primary coil due to the changing current in the secondary coil is E_M = M di₂ dt = 3 2 = 6 V . Since this induced emf opposes the current i₁ (which flows from P to Q), it acts as a potential drop when traversing from P to Q. The potential drop across the resistor is i₁ R = 2 5 = 10 V . The KVL equation is: V_P - i₁ R - E_M = V_Q V_P - V_Q