Question

The circuit shown has been operating for a long time. The instant after the switch in the circuit labeled S is opened, what is the voltage across the inductor V_L and which labeled point (A or B) of the inductor is at a higher potential ?

Take R₁ = 4.0 Ω, R₂ = 8.0 Ω, and L = 2.5 H.

(a) V_L = 12V, point A is at the higher potential 

(b) V_L = 12V, point B is at the higher potential 

(c) V_L = 6V, point A is at the higher potential 

(d) V_L = 12V, point B is at the higher potential

Answer 1

Correct option: (d) V_L = 12V, point B is at the higher potential

Explanation:

Current in the inductor before opening ‘S’

Since current in inductor does not change instantly, therefore, just after opening ‘S’,

12 + V_L – IR₁ = 0 12 + V_L – (4.5) (4) = 0 V_L = 6V with ‘B’ at higher potential.