An inductor with an inductance of 5.0 H and negligible resistance is connected in series with a 100 resistor and a battery having an emf of 2.0 V . Determine the potential difference across the resistor 20 ms after switching on the circuit.

Step-by-step solution

The current in an L-R circuit during growth is given by I = I₀(1 - e^ -t/ ) , where I₀ = E R is the steady state current and = L R is the time constant. The potential difference across the resistor at time t is: V_R = I R = E(1 - e^ -t/ ) Given values: E = 2.0 V L = 5.0 H R = 100 t = 20 ms = 0.02 s The time constant is: = L R = 5.0 100 = 0.05 s Substituting these values into the equation for V_R : V_R = 2.0 (1 - e^ -0.02/0.05 ) V_R = 2.0 (1 - e^ -0.4 ) Using the value e^ -0.4 0.67 : V_R = 2.0 (1 - 0.67) V_R = 2.0 0