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 = 1.00 s .

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 as a function of time is I(t) = E R (1 - e^ -t/ ) . At t = 1.00 s , the ratio t = 1.00 0.02 = 50 . Since e⁻⁵⁰ 0 , the current has reached its steady-state value I = E R . The potential difference across the resistor is V_R = I R = ( E R ) R = E = 5.00 V .