Susan Brown

[Susan Brown]

Hi Mariah,

Don’t you also know the circuit current?

Here are the critical things to remember about a circuit with loads in series:

1. The current will be the same at every point in the circuit.
2. The current flowing through each load will produce a voltage drop across that load.
3. The sum of those voltage drops through all of the loads will equal the source voltage.

Look at the Ohm’s Law equations in Unit 3. If you know the circuit current and you know each load’s resistance, do you see how you could calculate the voltage drop across each load?

[Susan Brown]

As far as resistance numbers go, practically speaking there’s no difference between 18.75 and 18.86. We don’t need to be that precise when it comes to resistance.

As far as a reset goes – please check your email.

[Susan Brown]

The equation is Req = 1/(1/R1 + 1/R2).

In the example in the Unit 5 video, the resistances are 30 and 50.

You always calculate the stuff in parentheses first: 1/30 + 1/50 = 0.0333 + 0.02 = 0.0533

Then 1/0.053 = 18.75

[Susan Brown]

Hi Robert,

Don’t feel dumb! This is a common question. Check out these two previous forum topics and see if they help you. Feel free to ask more questions after looking over these:

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/calculating-equivalent-resistance/

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/equivalent-resistance-in-parallel-circuits/

[Susan Brown]

Yes, it’s purpose is to relieve gas pressure buildup above the diaphragm in case the diaphragm develops a leak.

[Susan Brown]

Hi Jason,

First of all, it would probably be helpful for you to rewatch the Half-splitting video, if you haven’t already, where we go over this.

But here’s some more info.

Question 16: we haven’t disconnected any wires, and we energize the circuit. Measuring voltage across a load is attempting to measure voltage drop. When current flows through a load, it will create a voltage drop. So if you measure zero voltage drop across a load, it means no current is flowing. This means the voltage we are measuring is potential voltage. Furthermore, the fact that we measure this 120vac wrt N at each terminal of the element also demonstrates that the element is not open – it is acting like an unbroken wire.

Does all of that make sense?

Question 17: the circuit is now half-split on the L2 side of the element. Yes, the measurements are wrt N. The side where we measure 0 vac wrt N is the side that has the open. In this case, it is the L1 side that has the various controls.

Let me know if this clears it up, or if you have follow-up questions.

[Susan Brown]

Yes!

[Susan Brown]

That’s correct.

[Susan Brown]

That’s correct!

[Susan Brown]

Hi Wyatt,

Think of it in terms of a failure sequence and also the temperatures at which the compartments normally operate.

A normal freezer is at around zero degrees, but customers usually won’t detect a problem in the freezer until it gets closer to the freezing/melting point (the temps will “seem okay” to the customer). However, a healthy fridge is at 34-40 deg, and food spoilage will start as low as 45-50 degrees. This is where we will get the phone call from the customer.

In the initial stages of a defrost system failure in a single evap unit, the system can usually still create enough cold air to keep the freezer relatively close to its normal range, but without adequate air flow / chilling (as the coils get increasingly frosted over), it won’t be able to transfer enough cold to the FF compartment. Left uncorrected, the freezer will begin to get significantly warmer as the frost continues to build.

Does that make sense? Let me know.

[Susan Brown]

exactly! Only the lightbulb will have voltage drop, since it is the only load in the circuit.

[Susan Brown]

#26: you said that the fuse is a switch, and that switches don’t drop voltage.

Remember – measuring voltage is measuring a *difference* in charge between two points. If a closed switch acts like a wire, there will be no measurable voltage because the charge is the same.

Voltage *drop* is only measured across a functioning load (when current is flowing through, causing the load to do work).

Fuses are not a load.

Does this clear it up? Please tell me more about what you are thinking. This can take some effort to really nail down!

#7 – yes, you are measuring potential voltage from L1, but with an open neutral the current is not able to flow.

[Susan Brown]

#26: the question is

For the light to be on, what is the correct voltage drop across the fuse?

If you measured voltage across a fuse in a normally functioning circuit, what should you measure?

#7: if the open is on the L1 side, how are you measuring 120vac? Remember – N is the reference point.

[Susan Brown]

Well, zero voltage drop across a known-good load, when you know there is some voltage present, means you have an open (no current is flowing).

You measure the full 120vac. So what does that tell you about which side of the circuit has the open?

[Susan Brown]

Oops – we were posting at the same time. I said you were correct about the fuse not dropping voltage – and that’s what we ask in the question.

The load in the circuit (the bulb) will drop the voltage. Make sense?

[Author]

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