An LR circuit includes an inductor of 500 mH , a resistor of 25.0 , and a 5.00 V emf connected in series. Determine the potential difference across the resistor at t = 100 ms .

Step-by-step solution

The time constant of the LR circuit is given by = L R . Substituting the given values: = 500 10⁻³ 25.0 = 0.02 s The current in the circuit at time t is: I(t) = I₀ (1 - e^ -t/ ) = V₀ R (1 - e^ -t/ ) The potential difference across the resistor is: V_R = I(t) R = V₀ (1 - e^ -t/ ) At t = 100 ms = 0.1 s : V_R = 5.00 (1 - e^ -0.1/0.02 ) = 5.00 (1 - e⁻⁵) Using e⁻⁵ 0.0067 : V_R = 5.00 (1 - 0.0067) = 5.00 0.9933 = 4.9665 V 4.97 V