In the circuit shown in the figure, the resistors have resistances R_1 = R_2 = 1
ID: 1434485 • Letter: I
Question
In the circuit shown in the figure, the resistors have resistances R_1 = R_2 = 10 ohm, the battery has a voltage epsilon = 6 V, and the inductor is ideal (i.e.. it has no internal resistance) with inductance L = 100 mil. The switch has been open for a long time and all currents are initially zero. At time t 0 the switch is closed. A "skeleton circuit" showing junctions and loops is also shown below the figure. Immediately after the switch is closed at time, what is the current through the inductor? Immediately after the switch is closed, what is the voltage across the inductor? Which end of the inductor, the top or the bottom, will be at the higher voltage? A long time after the switch is closed the circuit reaches a steady state (all currents and voltages reach a steady, constant value). What is the voltage across the inductor a long time after the switch is closed? What is the current through the inductor a long time after the switch is closed? In which direction is this current flowing?Explanation / Answer
a) The current through inductor immediately after the switch is closed is ZERO because the current in inductor cannot change instantaneously. Initially there was zero current earlier as there was open circuit and no potential across registor
b)since current through inductor is ignored,
Potential across R2 = 6V * 10/(10+10) = 3V potential is divided equally if two equal R are connected in series
Voltage across inductor = Potential across R2 = 3V Answer
Top end will have higher voltage as in the case of R2
c) after a long time, inductor acts like sorted wire, so all current (after R1) will go through inductor and zero current will go through R2. Since there is no current in R2,
vOLTAGE across R2= 0 X 10 = 0
Voltage across inductor =Voltage across R2 = 0
d) After a long time, no current will flow through R2, so forget there is R2, and inductor will act like a sorted wire
So current through inductor = current in the circuit ignoring R2 and treating induactor as sorted wire
= 6 / 10 A = 0.6A Answer
The current will flow in clockwise i.e top to bottom in inductor