Susan Brown

[Susan Brown]

Hi Ted – I can set you back to Unit 4. FYI, it will clear the two quizzes you took after that, so you’ll need to retake them. Make notes if you want to, then let me know when you’re ready for the reset.

FYI, for future reference, there is a Quiz & Exam Reset Request form in the Campus Support menu. That’s the best way to get a reset.

[Susan Brown]

Sorry – I’m not sure what you mean. What would be 32?

The total resistance of the circuit is 32 ohms.

Did you rewatch the videos in Unit 4? There’s one where I do a some different calculations on loads in series, including the heat generated. And then there’s another where Samurai is finding the heat generated by a loose connection (with slightly different numbers for the resistances)

[Susan Brown]

To just find the heat generated by the 4-ohm loose connection, it would have to be [the voltage dropped across the loose connection] squared / 4.

You can do it that way, but I think it’s easier to find the current in the circuit, then use I squared x 4.

[Susan Brown]

P = E squared / R is a correct formula, but if you are going to use this, E has to be the voltage dropped across the R that you are interested in. Since there are two resistances in this circuit, you can’t use 240 for E and 4 ohms for R. Does that make sense?

We suggest using P = I squared x R, and step through a similar calculation in Unit 4 of Basic Electricity.

[Susan Brown]

To measure pressures, you have to tap into the sealed system – that’s what we mean by “invasive”. Temperature and amps don’t require that.

[Susan Brown]

Hi Anthony,

This question is referring to the actual measurements we would make to evaluate the sealed system. The point of the non-invasive techniques are to find ways to know the pressure without directly measuring it, which can only be done invasively. The Danfoss app technique you refer to is taking the direct measurement of temperature then combining that with thermodynamic knowledge to determine pressure.

It’s kind of like how you can look at a glass of ice water and know that its temperature is right around 32 degrees F. You didn’t actually stick a thermometer in it and directly measure it. But the visual information (ice in a glass with water) plus your knowledge of the characteristics of ice let you determine pretty accurately what the temperature is.

Does that make sense?

[Susan Brown]

Hi David,

Voltage drop is the voltage you measure across a working load. You can calculate the expected voltage drop using E = I x R or E = P/I, depending on what information you have.

So the amount of voltage drop will depend on the amount of current flowing through it. And this is influenced by the presence of any other loads in the circuit or things like variable speed systems.

The Samurai recommends that you watch these Workshops at Appliantology if you haven’t already to help you strengthen your understanding of this important topic!

Voltage, Voltage Drop, Loads & Switches, Jumpers & Cheaters

Schematic Analysis Using Ohm’s Law

Voltage Measurements and More

• This reply was modified 1 year, 9 months ago by Susan Brown.

[Susan Brown]

Correct, so the closed switch is a shunt. It bypasses the heater, but current is still going through the bulb, which means it is not a short.

[Susan Brown]

Doesn’t the current still have to go through the light bulb?

[Susan Brown]

And the switch let’s current get to the bulb without it going through the heater also

That is correct.

A shunt or short I have trouble telling them apart

A short allows current to go from Line to N without going through another load. Is that the case here?

[Susan Brown]

Wouldn’t the bulb get brighter with the switch closed

Yes – and why is that?

And the switch not matter when it comes to the heater heating

The switch does have an impact on the heater. Remember, a closed switch doesn’t have any resistance. When the switch is closed, it creates a special condition that we talked about in this lesson. Do you have any idea what that is?

[Susan Brown]

Can you describe how the current flows in A and B? What changes when that switch is closed in B?

[Susan Brown]

Hi Cooper,

We are talking about this diagram, correct?

[Susan Brown]

Oh – I think I figured it out. Mod. 4, unit 1, correct?

If you use E in the calculation of P, it has to be the voltage dropped across the load in question, not the source voltage. You could find the voltage drop across the element, but it’s usually easier to find the current instead, then use I^2 x R.

This is something we spend some time on in the Core course. Do you know the distinction between voltage and voltage drop in the context of an appliance circuit?

[Susan Brown]

Hi John,
Which Module and Unit is the video in?

[Author]

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