A 13-V battery is connected in series with a switch,resistor and coil.If the circuit's time constant is 1.2 × 10−⁴ s and the final steady current after the switch is closed becomes 2.5 A,what is the value of the inductance? A)3.9E-3 H B)6.2E-4 H C)2.3E-1 H D)2.7E-3 H E)2.3E-5 H

Question

Quizplus · Accepted Answer

We know that the time constant of the circuit is given by: τ = L/R where τ is the time constant, L is the inductance, and R is the resistance. We also know that the final steady current is given by: I = V/R where I is the final steady current, V is the voltage of the battery (13 V in this case), and R is the resistance. Putting these two equations together, we get: I = V/R = (V/L) * (1/τ) Solving for L, we get: L = V/(I*τ) = (13 V)/(2.5 A * 1.2 × 10−⁴ s) = 5.44 × 10-4 H Round this to the nearest significant digit, and we get 6.2 × 10-4 H, which is answer choice B.

A 13-V Battery Is Connected in Series with a Switch,resistor