Resistor and Current Source in Parallel with Short Circuit?

[BlueSand]

This circuit is really confusing me. I'm trying to find i using KVL and KCL. What's throwing me off is the short circuit effectively created by the inductor. I know that the whole circuit can't be short circuited because there is a current source that is preventing all of i from going through the inductor-branch. Is current going through the resistor as well, now? I need to figure out what the current is supposed to be doing here.

Here's what I have:

i +2i = 3i (in the wire leading to the inductor and resistor in parallel) This current will all go through the short circuit and none through the 200ohm resistor. But here it's odd because according to KVL, if I choose the loop to be through the 10V source, the 625 resistor and then the short circuit, the current i is 10/625 A. BUT doesn't the current source have to have the same voltage drop as the short circuit since they are in parallel? And how can a short circuit have any voltage drop at all? I have no idea.

 

[Antiphon]

You are right on all counts. Not sure what you're confused about.

 

[vk6kro]

A current source can deliver a current into a short circuit even though this means it will have zero volts across it.

Your diagram shows uF as the unit for inductance. It could probably mean uH.

This assumes the inductor has no resistance and the supply is pure DC.

 

[BlueSand]

Okay, I understand. I didn't realize that was possible, haha. Thanks!

 

[alphysicist]

I can understand your confusion with this circuit. Let me try to break it down for you.

First, let's clarify what a short circuit is. A short circuit is a low-resistance connection between two points in a circuit, which allows current to flow without any resistance. In this case, the inductor and the current source are in parallel, creating a short circuit.

Now, let's consider Kirchhoff's Voltage Law (KVL). KVL states that the sum of all voltage drops in a closed loop must equal the sum of all voltage sources in that loop. In this circuit, if we choose the loop to be through the 10V source, the 625 ohm resistor, and the short circuit, the voltage drop across the short circuit must be 10V, as this is the only voltage source in the loop.

But how can a short circuit have a voltage drop? Well, in this case, the short circuit is effectively acting as a wire with zero resistance. So, the voltage drop across it will be zero. This means that the voltage drop across the 625 ohm resistor must be 10V, as KVL states.

Now, let's consider Kirchhoff's Current Law (KCL). KCL states that the sum of all currents entering a node must equal the sum of all currents leaving that node. In this circuit, the current leaving the 10V source must equal the sum of the currents entering the node where the inductor and current source are connected. This means that the current through the 625 ohm resistor must be equal to the current through the inductor and current source.

So, to answer your question, yes, the current will flow through the resistor as well. In fact, the current through the resistor and the current through the inductor and current source must be equal. This is because the current source is providing a constant current, and KCL states that the sum of all currents entering a node must equal the sum of all currents leaving that node.

I hope this helps clarify things for you. Remember, when dealing with circuits, it's important to keep in mind the laws of Kirchhoff and to carefully consider the behavior of each component in the circuit.