Part A For the circuit shown in the figure, the inductors have no appreciable re
ID: 1913035 • Letter: P
Question
Part A For the circuit shown in the figure, the inductors have no appreciable resistance and the switch has been open for a very long time. http://session.masteringphysics.com/problemAsset/1543240/1/p24.43.jpg (a) The instant after closing the switch, what is the current through the 60.0- resistor? Part B (b) The instant after closing the switch, what is the potential difference across the 15.0-mH inductor? Part C (c) After the switch has been closed and left closed for a very long time, what is the potential drop across the 60.0- resistor?Explanation / Answer
a) at the instant when switch is closed , inductors behave as open circit I = 100/(10+60) A I = 10/7 A = 1.428571 A b) Since inductor behaves like an open circuit in the start, no current flows through it Pot. Diff. across 15 mH inductor = Pot. Diff. across 60 Ohm resistence Pot. Diff. across 15 mH inductor = I*r = (10/7)*60 V = 600/7 V = 85.714 V c) After long time inductor behaves as short circuit So, the 40 mH inductor is shorting the 60 Ohm resisor => Voltage across 60 Ohm resistnce = 0