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I want to check my answers for this problem: Part c) below doesn't agree with the book. And even though part a) agrees, I'm not sure about my method.
I've attached a figure for reference.
For the attached figure, the switch is closed fo t<0, and steadystate conditions are established. The switch is thrown open at t = 0.
a) Find the initial voltage across L just after t = 0. Which end of the coil is at higher potential: A or B?
c) How long after t = 0 does the current in the 6kOhm resistor have the value 2.00 mA.
For part a), I used Kirchhoff's loop rule to get the currents at steady state.
6kOhmI1 = 18V
I1 = 3mA
2kOhmI2  0.4H(di2/dt) = 18V ==> di2/dt = 0 at steady state.
I2 = 9ma
Itotal = 12mA
Once the switch is thrown open I have an RL series circuit.
In a series circuit I should be the same across both resistors so:
I1 goes to zero, so I total in the circuit is 9mA.
6kOhmI + 2kOhmI  0.4H(di/dt) = 0
0.4(di/dt) = 8kOhmI = 8kOhm(9mA) = 72Volts; B has the higher potential.
If my reasoning above is correct then for part c)
I = Iinitial(e^(t/T)) where T = L/R = .4/8000 = 50 microseconds
2mA = 9mA(e^(t/50us)
ln(1/6) = t/50us
t = 50ln(1/6) = 75.2 microseconds. The answer is the book is 75.2us
Thanks in advance for the help.
I've attached a figure for reference.
For the attached figure, the switch is closed fo t<0, and steadystate conditions are established. The switch is thrown open at t = 0.
a) Find the initial voltage across L just after t = 0. Which end of the coil is at higher potential: A or B?
c) How long after t = 0 does the current in the 6kOhm resistor have the value 2.00 mA.
For part a), I used Kirchhoff's loop rule to get the currents at steady state.
6kOhmI1 = 18V
I1 = 3mA
2kOhmI2  0.4H(di2/dt) = 18V ==> di2/dt = 0 at steady state.
I2 = 9ma
Itotal = 12mA
Once the switch is thrown open I have an RL series circuit.
In a series circuit I should be the same across both resistors so:
I1 goes to zero, so I total in the circuit is 9mA.
6kOhmI + 2kOhmI  0.4H(di/dt) = 0
0.4(di/dt) = 8kOhmI = 8kOhm(9mA) = 72Volts; B has the higher potential.
If my reasoning above is correct then for part c)
I = Iinitial(e^(t/T)) where T = L/R = .4/8000 = 50 microseconds
2mA = 9mA(e^(t/50us)
ln(1/6) = t/50us
t = 50ln(1/6) = 75.2 microseconds. The answer is the book is 75.2us
Thanks in advance for the help.
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