A series RL circuit is connected to an emf source of angular frequency $ \omega $. The current: A) leads the applied emf by tan^-¹( $ \omega $L/R) B) lags the applied emf by tan^-¹( $ \omega $L/R) C) lags the applied emf by tan^-¹( $ \omega $R/L) D) leads the applied emf by tan^-¹( $ \omega $R/L) E) is zero

Question

Quizplus · Accepted Answer

In a series RL circuit, the current lags behind the voltage due to the inductive reactance. The phase difference between the current and voltage is given by the formula tan⁻¹(XL/R), where XL is the inductive reactance and R is the resistance. In this case, XL = ωL, where L is the inductance of the circuit. Therefore, the phase difference is equal to tan⁻¹(ωL/R). Since the emf source is given by ω, the phase difference is equal to tan⁻¹(ωL/R) - π/2. Simplifying this expression gives tan⁻¹(Lω/R) - π/2, which is option (B).

A Series RL Circuit Is Connected to an Emf Source ω \omega ω

A Series RL Circuit Is Connected to an Emf Source $\omega$