In the given circuit the sliding contact is pulled outwards such that electric current in the circuit changes at the rate of 8 A / s. At an instant when R is 12 Ω, the value of the current in the circuit will be ________ A.

Solution

Physics sir · 3 steps

Pulling the contact out raises R, so the current is falling, and the inductor pushes to keep it going.

1. Sign of the change. Pulling the sliding contact outwards puts more of the resistor in the circuit, so R increases and the current decreases. So dI/dt = -8 A s⁻¹.
2. Induced emf. The inductor opposes the fall in current, so its emf acts in the same direction as the battery. Its size is L|dI/dt| = 3 × 8 = 24 V.

   E - LdI/dt = IR ⇒ 12 - 3 × (-8) = I × 12

3. So. 36 = 12I, which gives I = 3 A.

The common slip: 1. Ignoring the inductor and using I = 12/12 gives 1 A. The 24 V from the inductor adds to the battery while the current falls.

The idea to keep. In a circuit with an inductor, Ohm's law alone is not enough while the current changes. Use E - LdI/dt = IR, and decide first whether the current is rising or falling.

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