advertisement

Review of LR circuits This handout covers current and voltage changes in a single-loop circuit with a resistor and an inductor. For the circuit above the initial current is zero and the switch is closed at time t = 0+. We apply Kirchoffâ€›s Law, i.e., the sum of voltage drops around a loop is zero, and get: 0 = -VB + IR + L dI dt . A little rearranging to place each term in the form of a current gives: dI dt . VB = I + R b The solution of this equation - recall that a yourself – is: I(t) = VB 1 - e- t R ( ) so that I(t = 0+) = 0 and I(t VL(t) = L dx = log e (b) - log e (a) and try to solve it x ) = VB/R. The voltage drop across the inductor is: dI = VB d L 1 - e- t R dt dt ( e ) = LV R B or: VL(t) = VBe- t so that VL(t = 0+) = VB and VL(t ) = 0. -t = LVB R - t e R L