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enter image description here

Can you just explain to me the logic of this question because I am a bit confused here.

Why do we have a 120 volt and 30-volt value? My take out is this: there is a current flowing through 120-ohm resistor but the voltage across the resistor is 30 volt. Should I put another resistor to complete 120 volt? I tried to do it like that and parted 1A current like 0.25A and 0.75A but I got nowhere with this. And it doesn't make sense because I feel like I should use the 1A current to make Norton circuit. I am not trying to make you solve this question entirely but if you can help me out to understand this question a bit I would really appreciate it.

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    \$\begingroup\$ There's no "120V" component. That's 120 ohm load. \$\endgroup\$ – Kyle B Dec 22 '20 at 23:39
  • \$\begingroup\$ sorry, I assumed because of 120ohm x 1A \$\endgroup\$ – stevenwilson Dec 22 '20 at 23:40
  • \$\begingroup\$ That's why you're having a problem... You're not understanding what the question is really asking. It's seeing if you can do a Norton equivalent circuit conversion. Have you learned that in school yet??? \$\endgroup\$ – Kyle B Dec 22 '20 at 23:41
  • \$\begingroup\$ Hang on, I'll sketch something in an answer..... \$\endgroup\$ – Kyle B Dec 22 '20 at 23:41
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    \$\begingroup\$ What do you understand is the definition of a Norton circuit? Add that very simple schematic,( fill in any component values you know, leave any you don't) to the question if you can, using the built in schematic editor. \$\endgroup\$ – user_1818839 Dec 22 '20 at 23:58
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A voltage power actually consists of two pieces... An ideal voltage source (V1) and a series resistance (impedance) (R1). These values are not given to you. The question is asking you to find them.

The question states there's 1A flowing when the output is shorted (ideal ammeter)

So you have to ask yourself, what value resistor & voltage source will give 1A in this situation:

schematic

simulate this circuit – Schematic created using CircuitLab

and 30V in this situation:

schematic

simulate this circuit

What's above here is the "Thevenin" circuit. They're actually asking for a Norton equivalent. I find it much easier to work with Thevenin, but they're easily converted.

We can see in the 2nd circuit, the voltage across 120 ohms is 30V. The current therefore has to be 30/120= 0.25A. However, that's just a bonus .. We don't need this value to find the final answer. I'm just showing how such things are inferred.

So now we can have two equations we can use

First circuit (using ohms law) V1=I1*R1 We know I1=1 Amp

Second circuit (Using resistor divider rule) VM1=V1*(R2/(R1+R2)) We know R2=120 and VM1=30

Substituting...

Eq1: V1=R1 (because I1=1)

Eq2: 30=R1 * (120/(R1+120)) (Substitute R1 for V1...)

Solve for R1 = 40 ohms

As we determined previously, R1=V1, so V1 =40V

This is the THEVENIN circuit then:

schematic

simulate this circuit

You need to find the NORTON equivalent next.

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  • \$\begingroup\$ I see now so the norton equivalent should be a 1A current source with a 40 ohm resistor right? \$\endgroup\$ – stevenwilson Dec 23 '20 at 0:12
  • \$\begingroup\$ I felt enlightened. Thank you so much again I was SO confused. I understand it finally. I am relieved :D \$\endgroup\$ – stevenwilson Dec 23 '20 at 0:14
  • \$\begingroup\$ I misspoke .. You're right ;) Norton equivalent is a source of 1A and a parallel resistor of 40 ohms \$\endgroup\$ – Kyle B Dec 23 '20 at 0:22
  • \$\begingroup\$ Dude you are awesome! that was a huge help. \$\endgroup\$ – stevenwilson Dec 23 '20 at 0:23
  • \$\begingroup\$ No problem brother - Do a few more hundred of these problems and it'll be like riding a bike ;) \$\endgroup\$ – Kyle B Dec 23 '20 at 0:24
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One measurement is the current when the output is shorted. The other measurement is for voltage when the output has a load of 120\$\Omega\$. Does that help?

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  • \$\begingroup\$ An ideal ammeter is an equivalent to a short-circuit when connected across two terminals. \$\endgroup\$ – DKNguyen Dec 22 '20 at 23:37
  • \$\begingroup\$ Yes, that helps a lot! But I have a question how did we understand this. Is ammeter values always used in short circuits? \$\endgroup\$ – stevenwilson Dec 22 '20 at 23:41
  • \$\begingroup\$ This is an "ideal" ammeter, so it has no resistance. Having no resistance, when you connect it between two wires, you are short circuiting them. (A real ammeter has some resistance, but that is not what this problem asks). \$\endgroup\$ – Math Keeps Me Busy Dec 22 '20 at 23:43
  • \$\begingroup\$ Oh okay thank you so much! \$\endgroup\$ – stevenwilson Dec 22 '20 at 23:44

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