Given N,Area and length,calculate the inductance of the inductor:
L=uNA^2/L=7.6hernry
Given 18 gauge copper, find the resistance using the equation=pL/A
0.001 is diameter of the 18 gauge wire
0.05 is the radius of
Next, we calculate the time constant L/R=5*10^-6 s
Experiment:
1. Adjust the oscilloscope to square mode and function generator to 40kHz, Hit standby to see the graph.
2. Connect 150 ohm resistor, 110 V inductor, and oscilloscope in parallel. Find the frequency of the graph.
Time div=5*10^-6 second, and we estimate the unit of time is around 0.6.So our period is
0.6*(5*10^-6)=3.25*10^-6 s
1.
2.
Based on the graph, find the inductance and compare to the theoretical value.
Experimental value:
time constant=3.25*10^-6 s/In(2)=4.69*10^-6
L=R*time constant=150*4.69*10^-6=730 mH
Theoretical value:L=uNA^2/L=740 mH
We are introduced to RL circuit. RL ciucuit has a inductor and resistor in parallel. As the current in the circuit increases, the current flows through inductor increases exponentially to the maximum value
I(t)=Imax(1-e^(-t/timeconstant))
Time constant is L/R
solve a problem:
1A circuit contains a 35.0 mH inductor in series with a R1 = 120 ohm resistor, both of which are in
parallel with a R2 = 730 ohm resistor and a 45.0 V battery. The switch is closed for 170 micros and
then opened.
(A) What is the time constant when the switch is first closed?
(B) What is the value of the current in both resistors at the moment the switch is opened at 170 micros?
(C) What is the voltage drop across both resistors at 170 micros?
(D) How long will it take for the voltage drop across the inductor to be equal to 11.0 V after the switch is closed?
(E) What is the time constant of the circuit after the switch is open at 170 micros?
(F) How much energy is dissipated in the two resistors after the switch is open at 170 micros?
process of solving part D:
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