Susan Brown

[Susan Brown]

Hopefully you just misheard something we said.

I’m very glad to hear about the new job and that your being in our training helped! It doesn’t surprise me – we do have a good reputation! It’s too bad his other techs won’t take him up on the offer.

[Susan Brown]

but I had found a statement from the training that current was the same in parallel circuits

Can you tell me where this is in the course?

“current varies in each branch of a parallel circuit depending on the total resistance of loads in each corresponding branch of the parallel circuit. However, the current will be the same outside of the branches.”

Yep, that’s it. Also, the total current will equal the sum of the currents in the individual parallel circuits.

If you look over our quiz questions for Units 4 and 5, it’s pretty clear how current behaves in series and parallel circuits.

[Susan Brown]

Hi Sean,
Have you watched the video at the end of Unit 3 where we show how to do the calculation, using slightly difference numbers for resistance? Try imitating our calculation yourself on paper. If there’s a step you don’t quite follow, tell me what it is and I’m happy to help you further.

[Susan Brown]

You figured it out! And so glad to hear you’re enjoying the course so far – thanks for letting us know!

[Susan Brown]

Yes, that’s what it is good for!

[Susan Brown]

Current depends on the resistance in a circuit.

The only generalization you can make about parallel circuits is that they each have the same source voltage, since each branch is tied to Line and N (or L1 and L2 if a 240 circuit).

The current going through each parallel branch will depend on the resistance of the load(s) in the branch. Only if each branch had an identical load would the current would be the same.

Does that help?

[Susan Brown]

You got it! The key to this question is recognizing the shunt. Everything falls into place after that.

[Susan Brown]

When the detector switch is closed, they will all go through that circuit, since it has no load.

By definition, (which we gave in Unit 1 and also talked about in more detail in Unit 5) all the current will go through a shunt, if one is available. It’s how loads are controlled in a lot of circuit configurations – switches sometimes close to create a shunt that bypasses a load so that it won’t do any work.

An unintended shunt (that happens by accident) is a “short”.

[Susan Brown]

Hi Shawn,

A lot of people call those “finger drivers”. We got ours at our local hardware store, but here’s an example at Amazon:

https://amzn.to/2ZsUz3v

[Susan Brown]

Hi Bill,

I moved your question to a new topic, since it didn’t really have to do with that other thread.

See the video in this forum topic – it should help you!
https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/calculating-equivalent-resistance/

[Susan Brown]

There’s a big difference between a path with resistance (a load) in it and a path with none. A shunt is not like a parallel circuit, because ALL of the electrons will choose that path over one with a load in it.

Keeping that in mind, take another look at the circuits and see what you think.

[Susan Brown]

When electrons have two paths available to them, and one has a load and the other does not, what happens? (This is a particular circumstance that we address in Unit 5.)

[Susan Brown]

Forget about main for the moment.

If you are the igniter or the booster, don’t you have more than one option to get to N? Remember, electrons don’t “see” bends or connections in the wires.

[Susan Brown]

Hi Duane,

Thanks for posting a question! I did move it to the Basic Electricity forum, and gave it a new title.

Short answer to your question is no – there will never be a partial voltage drop across a load if it is the only load in the circuit.

It’s an observable law – “Kirchoff’s Law” – the sum of the voltage drops in a series circuit shall always, everywhere, and forever add up to the supply voltage. It’s related to the conservation of energy, but there’s no need to try to understand it deeper than that. We just have to accept it!

In the video you referred to, if we changed the lightbulb to something with a different resistance, that would change the current flow in the circuit.

I = E/R

The voltage is determined by the power supply, and the resistance in the circuit is determined by the loads that are in the circuit. The current will be whatever it takes to drop all the voltage across the load(s) in the circuit.

Does all of that help?

[Susan Brown]

By the way, here’s a great webinar recording at Appliantology that would be helpful, if you haven’t seen it yet:

https://appliantology.org/topic/72423-voltage-voltage-drop-loads-switches-jumpers-cheaters/

Also, we’ve got some recent videos at our YouTube channel that show voltage measurements on various jobs.

https://www.youtube.com/user/SamuraiRepairman/videos

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