Second Order Circuits: KCL/KVL & Complimentary Soln

• Engineering
• Tekneek
In summary: Once you have that, you can solve for the steady state current (iL) through the inductor. When the switch is closed for "a very long time", the inductor and capacitor will look like they are short-circuiting. The steady state current through the inductor would be 0.015Amps.
Tekneek

Homework Statement

For second order circuits, do you apply kcl/kvl in the circuit when the switch is open or closed to find the differential equation for complimentary solution?

For example for the circuit attached (the circuit has been operating for a long time with switch closed prior to t=0), applying kcl at the node after adding the two resistors in the parallel:

iL + V(t)/0.6kohms + [V(t)-VC(t)]/3kohms = 0

Not sure what to do after this (lets say if i want to find current in inductor when t>0)

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When the switch is closed for "a long time" so that the circuit achieves a steady state, what will the inductor and capacitor "look like" electrically? From that you should be able to determine the steady state current through the inductor. This will determine the starting conditions for the inductor circuit at the moment the switch opens...

gneill said:
When the switch is closed for "a long time" so that the circuit achieves a steady state, what will the inductor and capacitor "look like" electrically? From that you should be able to determine the steady state current through the inductor. This will determine the starting conditions for the inductor circuit at the moment the switch opens...

After long time inductor looks like short circuit and capacitor looks like open circuit. The steady state current through the inductor would be 0.015Amps. What i am trying to find is the equation for iL when t>0

Tekneek said:
After long time inductor looks like short circuit and capacitor looks like open circuit. The steady state current through the inductor would be 0.015Amps. What i am trying to find is the equation for iL when t>0

Right. So now you know the initial current that is flowing through the inductor when the switch opens. That's an initial condition. You should be able to write the differential equation by applying KVL.

gneill said:
You should be able to write the differential equation by applying KVL.

So do i apply the KVL assuming the switch is open or closed? Because the two would be different.

Tekneek said:
So do i apply the KVL assuming the switch is open or closed? Because the two would be different.

The switch opens at t = 0 and you want the circuit behavior for t > 0... surely that means you want the circuit behavior for when the switch is open.

You analyze the circuit when the switch is closed in order to find the initial conditions that start things off for the open switch arrangement.

Tekneek said:
So do i apply the KVL assuming the switch is open or closed? Because the two would be different.
You are looking for the differential equation of circuit behaviour when the switch is open.

Or more precisely, you are looking for the differential equation of circuit behaviour when the switch is opened and kept open.

gneill said:
The switch opens at t = 0 and you want the circuit behavior for t > 0... surely that means you want the circuit behavior for when the switch is open.

You analyze the circuit when the switch is closed in order to find the initial conditions that start things off for the open switch arrangement.

This means I will have two differential equations?

1) L(di(t)/dt) + i(t)R = 10

2) Cdv(t)/dt + V(t)/R = 0, where R is the two resistors in the Right of the switch added in series.

Solving 1.

General Solution: Ae^(-t/$\tau$) + B

Complimentary Solution will be Ae^(-t/$\tau$)

For Particular solution, iL(∞) = 10/1000 = 10mA = B

But when I try to solve just using differential equation I don't get B = 10mA

For example since my forcing function is a constant, I plug in a constant B in my diff eq, then I get

B = 10*(L/R)

Am solving my Diff. Eq. wrong way?

Tekneek said:
This means I will have two differential equations?

1) L(di(t)/dt) + i(t)R = 10

2) Cdv(t)/dt + V(t)/R = 0, where R is the two resistors in the Right of the switch added in series.

Solving 1.

General Solution: Ae^(-t/$\tau$) + B

Complimentary Solution will be Ae^(-t/$\tau$)

For Particular solution, iL(∞) = 10/1000 = 10mA = B

But when I try to solve just using differential equation I don't get B = 10mA

For example since my forcing function is a constant, I plug in a constant B in my diff eq, then I get

B = 10*(L/R)

Am solving my Diff. Eq. wrong way?

Your equation (1) looks like it pertains to the timeframe t > 0, when the switch is open. As such there should be the initial conditions that are in force at time t = 0. Those initial conditions are the result of the circuit activity that holds while the switch has been closed "for a very long time". Find those initial conditions first (that is, determine the current flowing through L at that time).

Your value for B should be the "final" current that flows through the inductor a long time after the switch has been opened. If you look a the terms of your differential equation, the component ##A e^{-t/\tau}## goes to zero as t → ∞, leaving just the B term.

gneill said:
Your value for B should be the "final" current that flows through the inductor a long time after the switch has been opened. If you look a the terms of your differential equation, the component ##A e^{-t/\tau}## goes to zero as t → ∞, leaving just the B term.

That is what i did and got B = 10mA. But shouldn't I get the same answer by just solving the Original (1) differential equation?

Tekneek said:
That is what i did and got B = 10mA. But shouldn't I get the same answer by just solving the Original (1) differential equation?

Sure. Can you show your work? It seems to me that when dI/dt goes to zero (i.e. steady state), you're left with an expression that should yield your value for B...

gneill said:
Sure. Can you show your work? It seems to me that when dI/dt goes to zero (i.e. steady state), you're left with an expression that should yield your value for B...

So my diff. eq. is

L(di(t)/dt) + i(t)R = 10

To find complimentary solution I make my diff eq. homogeneous (set right hand side to 0)
This gives me Ae..part which I understand.

But for particular solution, I look at the forcing function f(t) and use a trial function to solve the diff. eq.
And since my f(t)=10=constant I plug in let's say X (a constant) in my differential equation,

LdX/dt + XR = 10
0 + XR = 10
X=10/R = 10/1000 = 10mA

lol nvm i think i made a mistake earlier in my calculation they do turn out to be the same.

So as a general rule when you write a differential equation for these circuits with switch...it depends whether the switch is open or closed after a t=0?

Tekneek said:
So my diff. eq. is

L(di(t)/dt) + i(t)R = 10

To find complimentary solution I make my diff eq. homogeneous (set right hand side to 0)
This gives me Ae..part which I understand.

But for particular solution, I look at the forcing function f(t) and use a trial function to solve the diff. eq.
And since my f(t)=10=constant I plug in let's say X (a constant) in my differential equation,

LdX/dt + XR = 10
0 + XR = 10
X=10/R = 10/1000 = 10mA

lol nvm i think i made a mistake earlier in my calculation they do turn out to be the same.
That's good news!

So as a general rule when you write a differential equation for these circuits with switch...it depends whether the switch is open or closed after a t=0?

The general rule would be to determine the state of the circuit immediately before the switch changes (currents, potentials), then carry those over as the initial conditions for the circuit that "appears" after the switch changes. Usually that means finding the steady state conditions for current and potential differences ("after a long time...") for the time immediately before the switch change.

When dealing with first order situations (RC or RL circuits), often one can write out the expression for the desired current or voltage simply by knowing the initial and final conditions and "connecting" them by a suitable exponential function with the obvious time constant (RC or L/R). No solving of differential equations required

1 person
gneill said:
That's good news!

The general rule would be to determine the state of the circuit immediately before the switch changes (currents, potentials), then carry those over as the initial conditions for the circuit that "appears" after the switch changes. Usually that means finding the steady state conditions for current and potential differences ("after a long time...") for the time immediately before the switch change.

When dealing with first order situations (RC or RL circuits), often one can write out the expression for the desired current or voltage simply by knowing the initial and final conditions and "connecting" them by a suitable exponential function with the obvious time constant (RC or L/R). No solving of differential equations required

Great! Thanks!

1. What is a second order circuit?

A second order circuit is an electrical circuit that contains two energy storage elements, such as capacitors and inductors. These circuits are characterized by the presence of second-order differential equations in their analysis and have a more complex response compared to first order circuits.

2. What is KCL and KVL in relation to second order circuits?

KCL (Kirchhoff's Current Law) and KVL (Kirchhoff's Voltage Law) are fundamental principles used in the analysis of second order circuits. KCL states that the sum of currents entering a node must equal the sum of currents leaving the node, while KVL states that the sum of voltages around a closed loop must equal zero.

3. How do I apply KCL and KVL to solve second order circuits?

To apply KCL and KVL to solve second order circuits, you need to first draw the circuit diagram and label all the components. Then, write out the KCL and KVL equations for each node and loop, respectively. Finally, solve the resulting equations to find the unknown voltages and currents in the circuit.

4. What is the complementary solution in second order circuits?

The complementary solution in second order circuits refers to the solution of the differential equation that describes the behavior of the circuit when there are no external sources present. This solution is determined by the initial conditions (e.g. initial voltages and currents) of the circuit and represents the natural response of the circuit.

5. How do I determine the complete solution in second order circuits?

The complete solution in second order circuits is the sum of the complementary solution and the particular solution, which is the solution that takes into account the effect of external sources in the circuit. The particular solution can be found by solving the differential equation with the external sources included, while the complementary solution can be determined using the initial conditions of the circuit.

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