A coil with an inductance of 2.0 H and a resistance of 20 is connected to a battery having an emf of 4.0 V . Determine the current at the instant 0.20 s after the connection is made.

Step-by-step solution

The growth of current in an L-R circuit is given by the equation: I = I₀ (1 - e^ -t/ ) The steady-state current I₀ is: I₀ = E R = 4.0 20 = 0.20 A The time constant is: = L R = 2.0 20 = 0.10 s Substituting the values at t = 0.20 s : I = 0.20 (1 - e^ -0.20/0.10 ) I = 0.20 (1 - e⁻²) Using e⁻² 0.135 : I = 0.20 (1 - 0.135) = 0.20 0.865 = 0.173 A Rounding to two significant figures, the current is 0.17 A .