How to Determine iL(t) in a Series RLC Circuit with Switching?

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In summary, the switch closes and causes the current in the inductor to decrease by 18V/6ohm. The current in the capacitor also decreases by 3A, or 15V. The current in the series resistor (7ohm) also decreases by 1.75A.
  • #1
asdf12312
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Homework Statement


The switch is initially open, closes at t = 0, and then reopens at t = 0.5. Find iL(t).
swcu9c.png


Homework Equations


The Attempt at a Solution



i really have no idea how to do this. i have a series RLC circuit at t=0-, t=0.5 and the circuit with the 18V source at t=0, t=0.5-.

I got iL(0)=iL(0-)=0
Vc(0)=Vc(0-)=0

iL(0.5)=iL(0.5-)=18V/6ohm= 3A
Vc(0.5)=Vc(0.5-)=5ohm*3A= 15V
iL'(0.5)=vL(0.5)/L=15/2 = 7.5 A/s

a = R/2L (series)= 7ohm/4 = 1.75
w0 = LC^(-1/2)=1/sqrt(2/17)=2.92

underdamped because a<w0.

so far is this right? and also i have no idea how to find a single equation iL(t) for t=0-, t=1.5-, and also t=infinity. what do i do??
 
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  • #2
asdf12312 said:

Homework Statement


The switch is initially open, closes at t = 0, and then reopens at t = 0.5. Find iL(t).
swcu9c.png


Homework Equations


The Attempt at a Solution



i really have no idea how to do this. i have a series RLC circuit at t=0-, t=0.5 and the circuit with the 18V source at t=0, t=0.5-.

I got iL(0)=iL(0-)=0
Vc(0)=Vc(0-)=0

iL(0.5)=iL(0.5-)=18V/6ohm= 3A
Vc(0.5)=Vc(0.5-)=5ohm*3A= 15V
iL'(0.5)=vL(0.5)/L=15/2 = 7.5 A/s

a = R/2L (series)= 7ohm/4 = 1.75
w0 = LC^(-1/2)=1/sqrt(2/17)=2.92

underdamped because a<w0.

so far is this right? and also i have no idea how to find a single equation iL(t) for t=0-, t=1.5-, and also t=infinity. what do i do??

You are correct that the currents and voltages in the right-hand circuit (past the switch) are zero for t=0-.

But to solve for the behavior after the switch closes at t=0, you need to write the KCL equation for the currents leaving the top right node.

What is the relationship between voltage and current for an inductor? How about for a capacitor? You will use those relations when writing the KCL equation.
 
  • #3
1. Use KVL around the outer loop, writing everything in terms of I1, I2, and Vc.
2. Find I2 in terms of Vc'
3. Substitute I2 into the KVL equation in step 1 to solve for I1 in terms of Vc and Vc'
4. Use KVL around left loop, writing everything in terms of Vc, Vc' and Vc''
Note: iL = I1 - I2
5. From step 4, group terms to get an equation in the form:
Vc'' + a Vc' + b Vc = c
6. Solve for Vc(t)
7. Solve for iL(t)
- Recall that I1 and I2 can both be represented in terms of Vc and Vc'. Use your
result from step 6 to help you determine iL(t)
8. Redraw the circuit for t > 0.5 s and solve (Now it's a series RLC circuit)
- The initial values for this circuit are iL (0.5s) and Vc (0.5s)

this is the hints our instructor gave, didn't think i would need it. i would like to try it the loop method bcause it seems easier to understand. but i am still having problems, on step #4. so far i have:

1. I1+2(I2)+Vc=18
2. I2=iL= C*Vc' = Vc'/17
3. I1 = 18-Vc-(2*Vc')/17
4. How do i find Vc"? for the left loop i get:
I1+5(I1-I2)=18
when i substitute I1 and iL I get:
18-Vc-(2*Vc')/17 + 5(iL)=18
-Vc-(2*Vc')/17 + 5(iL)=0
 
Last edited:
  • #4
asdf12312 said:
this is the hints our instructor gave, didn't think i would need it. i would like to try it the loop method bcause it seems easier to understand. but i am still having problems, on step #4. so far i have:

1. I1+2(I2)+Vc=18
2. I2=iL= C*Vc' = Vc'/17
3. I1 = 18-Vc-(2*Vc')/17
4. How do i find Vc"?

KVL equations use voltages across components as you go around each loop. It looks like you are writing a mix of currents and voltages above?
 
  • #5
i am sorry i got confused. left out the 1ohm i multiplied I1 by, so that would make it voltage.
1. 1ohm(I1)+2ohm(I2)+Vc=18
2. I2=iC= C*Vc' = Vc'/17
3. 1ohm(I1) = 18-Vc-(2*Vc')/17

so it is now a voltage equation. how would i find I1 alone in terms of Vc'?
 
  • #6
asdf12312 said:
i am sorry i got confused. left out the 1ohm i multiplied I1 by, so that would make it voltage.
1. 1ohm(I1)+2ohm(I2)+Vc=18
2. I2=iC= C*Vc' = Vc'/17
3. 1ohm(I1) = 18-Vc-(2*Vc')/17

so it is now a voltage equation. how would i find I1 alone in terms of Vc'?

You need to write both KVL equations before starting to solve for t=0+.

And you will only have the right-hand loop obviously after the switch is opened, with some initial conditions determined by how the inductor current and capacitor voltage ramp up for that first 0.5 seconds...
 
  • #7
alright, so KVL equations for outer and left loop, and leave out the right loop for now?

1. for outer loop: 1(I1)+2(I2)+Vc=18
2. using equation I=C*dv/dt, since I2 is same as current through capacitor i got I2=iC= C*Vc' = Vc'/17

also for left loop, the equation before substituting for I1/IL is: 1(I1)+5(I1-I2)=18
also noting that IL=I1-I2i think this is right so far, but the problem is i don't know what to do from here. specifically how to do step #3, can't find a way to relate I1 to Vc'
 
  • #8
OK, suppose i try node method and write KCL for top node:
-I1+IL+IC=0

in terms of Vc, -I1=(Vc-Vs)/(R1+R3) = (Vc-18)/3
also, iC= C(dv/dt)

so:
Vc/3 + IL + C(d(Vc)/dt) = 6

Since Vc is equal to voltage across 5ohm plus the V across inductor:
Vc = L(d(IL)/dt) + IL(5)

if i substitute this into the equation:

1/3[L(d(IL)/dt) + 5(IL)] + IL + C(d/dt)[L(d(IL)/dt) + 5(IL)] = 6

following the method that my book uses:
IL'' + [L+(R1*R3*C)/(R1*L*C)]IL' + [(R1+R3)/(R1*L*C)]IL = Vs/(R1*L*C)

where R1=3, R3=5, in this case. also L=2 and forgot to mention that C actually simplifies to 1/17 = 0.05882F so:a = (L+R1*R3*C)/(R1*L*C) = (2+(15/17)) / (6/17) = 8.1666
b = (R1+R3)/(R1*L*C) = 8/(6/17) = 22.666
c = Vs/(R1*L*C) = 18/(6/17)= 51

A= a/2 = 4.08
w0 = sqrt(b) = 4.76

A<w0 so underdamped.
 
Last edited:

FAQ: How to Determine iL(t) in a Series RLC Circuit with Switching?

1. What is the formula for determining iL(t) in a circuit?

The formula for determining iL(t) in a circuit is iL(t) = V/R * (1 - e^(-Rt/L)), where V is the voltage, R is the resistance, t is time, and L is the inductance.

2. How do I calculate the value of iL(t) at a specific time in the circuit?

To calculate the value of iL(t) at a specific time, plug in the values for V, R, L, and the desired time into the iL(t) formula and solve for iL(t).

3. Can the value of iL(t) change over time in a circuit?

Yes, the value of iL(t) can change over time in a circuit as it is influenced by the changing voltage and resistance in the circuit. This is especially true in circuits with inductors, as the inductor's magnetic field can store energy and cause changes in current over time.

4. How does the value of iL(t) relate to other variables in the circuit?

The value of iL(t) is directly related to the voltage and resistance in the circuit, as well as the inductance of any inductors present. As these variables change, the value of iL(t) will also change.

5. Are there any other factors that can affect the value of iL(t) in a circuit?

Yes, other factors such as the type of inductor used and the presence of other components in the circuit can also affect the value of iL(t). Additionally, external factors such as temperature and environmental conditions can also have an impact on the value of iL(t).

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