# Solving Partial Circuit: Determine Current as Function of Time

• Gear300
In summary, a partial circuit shown in the attachment was discussed. A current pulse of 10.0A was fed to the circuit, which then became zero again after 200 microseconds. The current in the inductor as a function of time was determined using Kirchoff's Method, and a differential equation was solved to get I3 = 10(1 - e^(-10000t). However, for t > 200 microseconds, the initial current is 63.9A instead of 0A. To find this, the equation -L*dI/dt - IR = 0 was used and solved to get I = I0*e^(-10000t). The initial current, I0, is not limited to
Gear300
Partial circuit shown in attachment. A current pulse is fed to the partial circuit shown in Figure P32.25. the current begins at zero, then becomes 10.0A between t = 0 and t = 200 microseconds, and then is zero once again. Determine the current in the inductor as a function of time. I1 is the current before the junction, I2 is the current through the resistor, I3 is the current through the inductor, and R is the 100 Ohm resistance.

The current is 0A when t < 0s. For t between 0 and 200 microseconds, I used Kirchoff's Method and came up with:
L*dI3/dt = (I1 - I3)*R...from here on I came up with a differential equation and got:
I3 = 10(1 - e^(-10000t), which is the answer I'm supposed to get, although when coming up this, I had to assume I1 was unaffected by the inductance and remained as 10A through the time interval.

Where I am stuck at is finding the current when t > 200 microseconds. The answer I'm supposed to get is I3 = (63.9A)*e^(-10000t), in which 63.9A is the initial current right after the current pulse is gone. I can get the format of the equation, but I just don't know how to find that the initial current is 63.9A...help?

#### Attachments

• PartialCircuit.png
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Not sure, but can you show us what you did to find the 'format of the equation'?

Are you sure you specified the integral limits correctly when solving the DE?

After the current pulse is gone, there should be an emf induced by the inductor that would provide for a current. I took the inductor-resistor loop and used Kirchoff's method to come up with:
-L*dI/dt - IR = 0...using that, I came up with I = Io*e^(-10000t), in which Io is the initial current.

63.9A is not the current after the current pulse is gone. you get the current after the
pulse is gone by substituting t = 200 microseconds in (63.9A)*e^(-10000t).

if you evaluate the first equation you had for 0 <= t <= 200 microseconds, you will
find that that will give the same value for I3 at t = 200 microseconds.

wait...can you expand on this (I'm not exactly getting it)?

if you have $$I = I_0 e ^ {\frac{- t}{R C}}$$ than I_0 is only the initial current, if you mean the current at t=0 with that. Here the initial current after the switch closes is

$$I = I_0 e ^ {\frac{- 0.0002}{R C}}$$

wait...doesn't RC refer to resistor-capacitor system? There are no capacitors in this partial circuit.

Gear300 said:
wait...doesn't RC refer to resistor-capacitor system? There are no capacitors in this partial circuit.

sorry. replace $\frac {-t}{R C}$ by $\frac { - R t} {L}$ in the previous post.

oh, ok...I think I get what you're saying; there is a functional continuity, in which at t = 200 microseconds for the first one and t = 0s for the second one come down to the same value, and I sort of realize my mistake in modeling the situation after the pulse in that I was assuming the resistor carried the same current the inductor did, although there is a junction where the current could split. So then, how did they come up with 63.9 or the equation in general?

How they came up with the equation in general is easy, you just repeat what you did for the first question (kirchhoff).

This yields: $$-iR -L \frac{di}{dt} = 0$$
Solving this yields: $$- \frac{R}{L} dt = \frac{di}{i}$$ and taking integrals:
$$\int_0^t - \frac{R}{L}dt = \int_{I_0}^i \frac{di}{i}$$.
Then after rearranging you get:
$$i = I_0 \exp (- \frac{Rt}{L} )$$

How you get the value of $$I_0$$ however I can't see so quickly... I'll think about it.

yeah...that value doesn't seem to come to me.

I did some thinking (just short) but I still don't know... anyone else maybe?

What I did think was weird though, if the current is actually jumping from 10A to 0A in an infinitely small amount of time (like the question implies), shouldn't the emf induced in L be infinite? Since the emf is $L \frac{di}{dt}$ and the rate of change in $i$ is infinite? Argh I don't know :p

heh...thats what I was sort of thinking. I typed the question and did the diagram exactly as they are. I'm not sure what's going on with this question.

The current through the inductor can not jump. It will be the same just before and just after 200 microseconds. that is how you find I_0.

So what you're saying is if I set the before and after equations equal, I could solve for I_0...but when I do that, the limit as to how much I_0 can go up to is 10A, isn't it?

Gear300 said:
So what you're saying is if I set the before and after equations equal, I could solve for I_0...but when I do that, the limit as to how much I_0 can go up to is 10A, isn't it?

The current through the inductor is limited to 10 A. that does not mean that I_0 is limited in the expression below. (I is smaller than 10 A if t>= 0.0002)

$$i = I_0 \exp (- \frac{Rt}{L} )$$

oh...I see what happened. I've always been taking t = 0s for that expression...a mistake I missed. Why is it .0002s?

## 1. What is a partial circuit and why is it important to solve?

A partial circuit is a portion of an electrical circuit that is not complete. It is important to solve because it can affect the overall functioning and safety of the entire circuit. Additionally, solving a partial circuit allows for accurate analysis and troubleshooting of the circuit.

## 2. How do you determine the current as a function of time in a partial circuit?

The current as a function of time in a partial circuit can be determined by using Kirchhoff's laws and Ohm's law. Kirchhoff's current law states that the sum of currents entering and exiting a node must be equal, while Kirchhoff's voltage law states that the sum of voltage drops in a closed loop must be equal to the applied voltage. By applying these laws and using Ohm's law (V=IR), the current can be calculated at different points in the circuit as it changes over time.

## 3. What factors can affect the current in a partial circuit?

The current in a partial circuit can be affected by several factors, including the resistance of the components in the circuit, the voltage applied, and the number of parallel and series connections in the circuit. Changes in any of these factors can cause the current to vary over time.

## 4. Can computer simulations be used to solve partial circuits?

Yes, computer simulations can be used to solve partial circuits. There are various software programs available that use mathematical models to simulate the behavior of a circuit and calculate the current as a function of time. These simulations can be useful in predicting the performance of a circuit and identifying potential issues before physically building the circuit.

## 5. How can solving a partial circuit be applied in real-world situations?

Solving a partial circuit can be applied in real-world situations in a variety of industries, such as electrical engineering, electronics, and renewable energy. It can be used to design and optimize circuits, troubleshoot and repair faulty circuits, and analyze the performance of complex systems. Additionally, understanding how to solve partial circuits can be beneficial for anyone working with electronics or electricity, as it provides a deeper understanding of how circuits work and how to ensure their safe and efficient operation.

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