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

In summary, current is flowing through the inductor, but the resistor doesn't have any current through it.
  • #1
KillerZ
116
0

Homework Statement



The switch has been closed for a long time.
169hnc1.jpg


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

ru4d8g.jpg


I used the KVL equation around the circuit to get:

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

I am not sure if this is the right way to do this.
 
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  • #2
-L *di/dl is the emf arising if the current change. So your equation is

[tex]
-0.1\frac{di_{L}}{dt} = 1i_{L}
[/tex]


The starting condition is [tex]
I_{L}(0)=1 A
[/tex]

The equation is easy to solve.

ehild
 
  • #3
Ok I attempted this and here is what I got:

KVL:

[tex]iR + v = 0[/tex]

[tex]iR + L\frac{di}{dt} = 0[/tex]

[tex]\frac{di}{dt} = -\frac{iR}{L}[/tex]

[tex]-\frac{di}{i} = \frac{R}{L}dt[/tex]

[tex]-lni + C = \frac{R}{L}t[/tex]

[tex]I_{L}(0)=1 A[/tex]

[tex]-ln1 + C = \frac{R}{L}0[/tex]

[tex]C = 0[/tex]

[tex]-lni + 0 = \frac{R}{L}t[/tex]

[tex]0 = \frac{R}{L}t + lni[/tex]

[tex]1 = e^{\frac{R}{L}t} + i[/tex]

[tex]i = 1 - e^{\frac{R}{L}t}[/tex]

[tex]v = L\frac{di}{dt}[/tex]

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

[tex]v = -Re^{\frac{R}{L}t}[/tex]
 
  • #4
KillerZ said:
[tex]0 = \frac{R}{L}t + lni[/tex]

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

KillerZ said:
[tex]1 = e^{\frac{R}{L}t} + i[/tex]

You have a "* " instead of "+"

[tex]i = e^{-\frac{R}{L}t}[/tex]

[tex]v = \frac{di}{dt}[/tex]

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

[tex]v = -Re^{-\frac{R}{L}t}[/tex]

ehild
 
  • #5
Ok I got it thanks for the help.
 

1. What is KVL and how does it apply to RL circuits?

KVL (Kirchoff's Voltage Law) states that the sum of all voltages in a closed loop must equal zero. In an RL circuit, this means that the sum of the voltage drops across the resistor and inductor must equal the total applied voltage.

2. How do I calculate the voltage across a resistor or inductor in an RL circuit?

To calculate the voltage across a resistor, you can use Ohm's Law (V=IR) where V is the voltage, I is the current, and R is the resistance. To calculate the voltage across an inductor, you can use the equation V=L di/dt where V is the voltage, L is the inductance, and di/dt is the rate of change of current.

3. How do I find the current in an RL circuit?

You can use Ohm's Law (I=V/R) to calculate the current in a resistor, where I is the current, V is the voltage, and R is the resistance. To calculate the current in an inductor, you can use the equation I=V/R where I is the current, V is the voltage, and R is the reactance of the inductor (XL=2πfL).

4. How does time affect an RL circuit?

Time affects an RL circuit through the rate of change of current (di/dt). As time increases, the rate of change of current decreases, which leads to a decrease in the voltage across the inductor. This is due to the inductor opposing changes in current flow, causing a delay in the circuit's response.

5. How do I solve an RL circuit homework problem?

To solve an RL circuit homework problem, you can follow these steps:

  • 1. Draw the circuit diagram and label all known values.
  • 2. Apply KVL to the closed loop to form an equation.
  • 3. Use Ohm's Law and/or the inductor equation to solve for any unknown values.
  • 4. Substitute the known values into the equation from step 2 and solve for the remaining unknown values.
  • 5. Check your answer by verifying that KVL holds true for the circuit.

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