Solve RL Circuit Homework: KVL, Voltage, Current, Time

[KillerZ]

Homework Statement

The switch has been closed for a long time.

a) Current flowing through inductor. Voltage across it. IS any current flowing through the resistor.

b) The switch opens at t = 0s. Find an equation for the inductor voltage as a function of time for t > 0. How long does it take for the inductor to reach 10% of its maximum value?

Homework Equations

Kirchhoff's Voltage Law (KVL)

The Attempt at a Solution

a) 1A is flowing throw through the inductor as the circuit is stable and the inductor is replaced with a short circuit this causes the resistor to have no current through it and the inductor to have no voltage across it.

b) This is where I am not sure about:

t > 0

I used the KVL equation around the circuit to get:

-(100)10^{-3}\frac{di_{L}}{dt} + 1i_{L} = 0

I am not sure if this is the right way to do this.

 

[ehild]

-L *di/dl is the emf arising if the current change. So your equation is

<br /> -0.1\frac{di_{L}}{dt} = 1i_{L} <br />


The starting condition is <br /> I_{L}(0)=1 A<br />

The equation is easy to solve.

ehild

 

[KillerZ]

Ok I attempted this and here is what I got:

KVL:

iR + v = 0

iR + L\frac{di}{dt} = 0

\frac{di}{dt} = -\frac{iR}{L}

-\frac{di}{i} = \frac{R}{L}dt

-lni + C = \frac{R}{L}t

I_{L}(0)=1 A

-ln1 + C = \frac{R}{L}0

C = 0

-lni + 0 = \frac{R}{L}t

0 = \frac{R}{L}t + lni

1 = e^{\frac{R}{L}t} + i

i = 1 - e^{\frac{R}{L}t}

v = L\frac{di}{dt}

v = L\frac{d(1 - e^{\frac{R}{L}t})}{dt}

v = -Re^{\frac{R}{L}t}

 

[ehild]

0 = \frac{R}{L}t + lni

Is is correct up to here, but wrong from here.

1 = e^{\frac{R}{L}t} + i

You have a "* " instead of "+"

i = e^{-\frac{R}{L}t}

v = \frac{di}{dt}

v = L\frac{d(e^{-\frac{R}{L}t})}{dt}

v = -Re^{-\frac{R}{L}t}

ehild

 

[KillerZ]

Ok I got it thanks for the help.