Help solving First Order RL circuit

[Chandasouk]

http://img713.imageshack.us/i/39642770.jpg

At time t = 0_

The circuit looks like the voltage source in series with R1 and R2 with the inductor acting like a short circuit.

iL (0_) = 60V/50Ω = 1.2A

At t = 0+

The circuit looks like

http://img839.imageshack.us/i/52735488.jpg

Would iL(0 +) be a different current or the same? I have a feeling it would be the same.

It is true that the inductance current remains the same right before and right after the switch is activated because the current through the inductor can't change instantaneously right?

Any help on solving fr iL( 0+) and iL(t) would be appreciated

 

[Metaleer]

Yes, it is true that the current cannot change instantaneously in a coil because consider the following equation,

[tex]u_L(t) = L \frac{di_L}{dt}.[/tex]​

Let's assume [tex]i_L(0^-) \neq i_L(0^+)[/tex], that is, [tex]i_L(t)[/tex] is not continuous at [tex]t = 0[/tex]. Then the derivative in the first equation is not bounded, and this means a voltage drop that goes to infinity, which is not physically possible.

I will now present to you an additional proof. Let us separate variables in the first equation and integrate; you will get this:

[tex]i_L(t) = i_L(t_0) + \frac{1}{L}\int_{t_0}^{t} u(\tau) d\tau.[/tex]​

Let us study what happens when [tex]t=0^+[/tex] and [tex]t_0 = 0^-[/tex]; we obtain

[tex]i_L(0^+) = i_L(0^-) + \frac{1}{L}\int_{0^-}^{0^+} u(\tau) d\tau.[/tex]​

Now, provided that the function [tex]u(\tau)[/tex] is bounded, it vanishes. This means that in a very idealized situation, when you have sharp voltage surges (mathematically idealized with what are known as Dirac delta "functions"), you could get a non-continuous function for the current in a coil. However, you don't need to worry, as this is a mathematical treatment of an idealized situation. In real life, you'll always have a small but non-zero resistance in a cable which is apparently a short circuit.

For [tex]i_L(t)[/tex], you'd need to get differential equations with Kirchhoff's laws and the relationship between voltage drop and current (see the first formula) in an inductor and then solve it, using the initial value you have determined.

Hope this helps. :)

 

[Chandasouk]

I try solving using node equations

I make a node called V1 where the three resistors intersect each other and another node called VL above the inductor

The node equation for V1 is

(V1-60)/30 + V1/60 + (V1-VL)/20 = 0

V1/30 - 2 + V1/60 + V1/20 - VL/20 = 0

The node equation for VL is

(VL-V1)/20 + 1.2 = 0

VL/20 = V1/20 - 1.2

VL = V1 -24

Plugging that back in the node equation for V1

V1/30 - 2 + V1/60 + V1/20 - (V1-24)/20 = 0

V1/30 - 2 + V1/60 + V1/20 - V1/20 + 24/20 = 0

V1 = -24

so VL = -24 - 24 = -48V ?

I have a feeling something is wrong.

 

[Metaleer]

You don't need to solve for the voltage drop in the inductor. Assume that for [tex]t \geq 0[/tex], the current in the branch to the left is [tex]i_1[/tex] going up, the branch in the middle has a current of [tex]i_2[/tex] going down, and the branch to the right has [tex]i_L[/tex] going down as already indicated in your diagram. We use Kirchhoff's voltage law for the left mesh:

[tex]V_s = R_1 i_1(t) + R_3 i_2(t)[/tex]​

and for the right mesh,

[tex]R_3 i_2(t) = R_2 i_L(t) + L \frac{d i_L}{dt}[/tex]​

and then apply Kirchhoff's junction rule to the node on top:

[tex]i_1(t) = i_2(t) + i_L(t).[/tex]​

Now, all you need to do is, for instance, plug the value of [tex]i_1(t)[/tex] from the last equation into the first equation, and from this new equation, solve for [tex]i_2(t)[/tex] and finally, plug this value of [tex]i_2(t)[/tex] into the equation in the middle. This will produce a first order linear differential equation in [tex]i_L(t)[/tex], where the initial value is none other than [tex]i_L(0^-) = i_L(0^+) = i_L(0)[/tex], which you will need to solve for the constant of integration that appears.

Once you have [tex]i_L(t)[/tex], all you need to do is differentiate it with respect to time and multiply it by [tex]L[/tex] to get the voltage drop in the coil as a function of time.

Good luck. ;)

 

[DeeNos]

.

I would first confirm that the circuit is indeed a first-order RL circuit, which means it has only one resistor (R1 and R2 are in series) and one inductor. If this is the case, then yes, the inductor current (iL) would remain the same before and after the switch is activated.

To solve for iL(0+), we can use the formula iL(0+) = iL(0-) + V/R, where iL(0-) is the current right before the switch is activated and V/R is the current flowing through the resistor after the switch is activated. In this case, V = 60V and R = 50Ω, so iL(0+) = 1.2A + (60V/50Ω) = 2.2A.

To solve for iL(t), we can use the formula iL(t) = iL(∞) + [iL(0+) - iL(∞)] * e^(-t/τ), where iL(∞) is the steady state current and τ = L/R is the time constant of the circuit. In this case, iL(∞) = 60V/50Ω = 1.2A and τ = 2H/50Ω = 0.04s. Plugging in these values, we get iL(t) = 1.2A + (2.2A - 1.2A) * e^(-t/0.04) = 1.2A + 1A * e^(-25t). This equation can be used to calculate the inductor current at any time t after the switch is activated.

It is important to note that the inductor current will eventually reach the steady state value of 1.2A, but it will take some time to do so. This is due to the inductor's property of opposing changes in current, which causes it to "store" energy and release it over time. Solving for the inductor current at different time points can help us understand the behavior of the circuit and its components.

I hope this helps in solving your first-order RL circuit. If you need further assistance, please don't hesitate to reach out to a qualified electrician or a physics tutor for additional support.