Don't box yourself into thinking you can only use parallel combinations. Do you see an easy way to use two resistors to get something that is a lot closer to 590 Ω than 614 Ω is?

But even 614 Ω is closer than is justified given that loading wasn't taken into account (beyond setting the current in the divider high enough so that it shouldn't be a big issue).

You might hedge your bets and come up with a set of resistor values on the assumption that you only have one of each kind of resistor available (which means that you need to assume that the resistors in the original circuit don't have to be constructed out of your set of resistors).

I think you are going to do great on this lab, particularly if you understand things well enough to start from scratch with a similar problem and work it through.

Hahah, of course a series of 500 ohms and 100 ohms would be a lot better!

Are you sure it is possible to achieve this task using only one of each resistor?

 

Hahah, of course a series of 500 ohms and 100 ohms would be a lot better!

Are you sure it is possible to achieve this task using only one of each resistor?

I don't know how close you can get, but the number of combinations you need to explore is actually pretty limited. You just list all of the possible values for R4 that you can make with a single resistor and probably with two resistors in either series or parallel and then calculate the values you need for the other three. Then choose combinations from your dwindling set of choices. A little bit of cleverness with a spreadsheet and you can probably work through them pretty quickly and just pick the one that comes closest. Then you have it in your hip pocket to pull out in case it turns out that you have to live with that restriction.

But I'm fairly confident (not certain, by any means) that you will have several of each value available.

 

When I first encountered that type of problem ... it appeared like this.

And there were questions ....





Source: Module one, Chapter Three, NEETS.

 

What was the "correct" answer to the first question?

I would say that it is a poorly worded question since R1 does not HAVE to be calculated first. We might CHOOSE to calculate it first for convenience, but convenience certainly doesn't dictate that we MUST calculate it first. Still, of the options offered, Option 1 is the best. A better wording of that option would have simply been. "It doesn't."

You can easily calculate the three (effective) resistor values independently once you decide how much excess current you want in your divider circuit. In this case they chose 5 mA and so you need 55 mA dropping 33 V across the top resistor (600 Ω), the middle one has to drop 30 V while carrying 15 mA (2 kΩ), and the bottom one has to drop 100 V while carrying 5 mA (20 kΩ).

EDIT: Changed 'does HAVE' to 'does not HAVE'.

Thanks, TheElectrician, for pointing out the typo.

 

3 was the correct answer. I can't take credit for writing the questions. It requires a full time position to deal with re-writing ambiguous questions in some books.

There was another one where you had to draw the circuit .... but that is too much like what the TS is doing right now. I just wanted to give them some ideas on where their question was headed.

I can remember a question on an FCC test where a question about bleeder resistors where the "best" answer was dealing with voltage regulation as safety was absent from all the answers.

Even though R1 in this circuit must be calculated first, it isn't a requisite for any circuit. Hence, 4 was a distractor with the "any circuit" descriptor.

If your like me, a speed reader at times, I miss words and must make a conscience effort to slow down and read all the answers after determining the stem of the question.

 

I still don't see that R1 has to be calculated first. I calculated it last above.

 

What was the "correct" answer to the first question?

I would say that it is a poorly worded question since R1 does not HAVE to be calculated first. We might CHOOSE to calculate it first for convenience, but convenience certainly doesn't dictate that we MUST calculate it first. Still, of the options offered, Option 1 is the best. A better wording of that option would have simply been. "It doesn't."

You can easily calculate the three (effective) resistor values independently once you decide how much excess current you want in your divider circuit. In this case they chose 5 mA and so you need 55 mA dropping 33 V across the top resistor (600 Ω), the middle one has to drop 30 V while carrying 15 mA (2 kΩ), and the bottom one has to drop 100 V while carrying 5 mA (20 kΩ).

Is there a missing word in the above, which I've inserted in red?

 

Yes. Thanks for spotting that.

It's amazing how often I either don't put in 'not' (or similar negation) when I meant to or do put it in when I didn't want to.

 

In this case they chose 5 mA

Yes they did. Typically, that problem has the bleeder unmarked and wants the students to determine both the current draw and the value of said resistor. The TS problem is an extension of that problem where they should have been able to extract the solution using associative learning, unless the previously learned material wasn't retained or learned.

 

Yes. Thanks for spotting that.

It's amazing how often I either don't put in 'not' (or similar negation) when I meant to or do put it in when I didn't want to.

Just wanted to let you know I built the circuit today (and after a little bit of troubleshooting), it worked! I had all the right voltage drops and rises! Thanks again for all your help

 

Just wanted to let you know I built the circuit today (and after a little bit of troubleshooting), it worked! I had all the right voltage drops and rises! Thanks again for all your help

Glad to hear it -- and glad to help.

When I get a chance I'll show how to solve it using the technique I was talking about a bit ago.

 

Hello, I was wondering how the OP has used kirchoff's law to calculate the current of 82.2 micro amps and 81.1 micro amps, what is the original voltage they are using from the the 30V possible Power supply? I tried calculating the voltage from the initial current of 0.0000163 amps across the total resistance of the circuit only to get 10.3V? Is this the voltage they must've used?

 

Hello, I was wondering how the OP has used kirchoff's law to calculate the current of 82.2 micro amps and 81.1 micro amps, what is the original voltage they are using from the the 30V possible Power supply? I tried calculating the voltage from the initial current of 0.0000163 amps across the total resistance of the circuit only to get 10.3V? Is this the voltage they must've used?

I don't know how you are getting 10.3 V. Please show your work so that I can spot what you are doing wrong.

The original circuit (solved) is shown in this post:

https://forum.allaboutcircuits.com/...with-three-outputs.134645/page-2#post-1123075

Ignore all of the white annotations and see if you can solve for the voltages and currents. Show your work and then I can look it over and make suggestions.

 

I don't know how you are getting 10.3 V. Please show your work so that I can spot what you are doing wrong.

The original circuit (solved) is shown in this post:

https://forum.allaboutcircuits.com/...with-three-outputs.134645/page-2#post-1123075

Ignore all of the white annotations and see if you can solve for the voltages and currents. Show your work and then I can look it over and make suggestions.

I just added the entire resistance in the circuit to which I got about 63333.333 ohms and i was trying to find out where OP got the current of 163 micro amps so I used V = IR thus = 0.000163 x 63333.333 which equaled about 10.3V for the current of the entire circuit. Also how would you calculate the current through each resistor, for example for the 10k ohm resistor, would it not be 8/10000 = 0.00008 amps or 800 micro amps but how does that make sense if the current in the enitre circuit is only 163 micro amps? PLease help I beg.

 

I just added the entire resistance in the circuit to which I got about 63333.333 ohms

The entire resistance of the circuit as seen between what two points.

and i was trying to find out where OP got the current of 163 micro amps so I used V = IR thus = 0.000163 x 63333.333 which equaled about 10.3V for the current of the entire circuit.

This is the voltage between what two points?

Also how would you calculate the current through each resistor, for example for the 10k ohm resistor, would it not be 8/10000 = 0.00008 amps or 800 micro amps but how does that make sense if the current in the enitre circuit is only 163 micro amps? PLease help I beg.

The reason it doesn't make sense is because you are abusing Ohm's Law. The formula V=IR relates the voltage ACROSS a resistor and the current THROUGH that SAME resistor to the resistance of THAT resistor. You are grabbing a handy voltage value that is NOT across the 10 kΩ resistor (the 8 V is the voltage across the 8 V source, which is NOT in parallel with the 10 kΩ resistor) and throwing at Ohm's Law.

There are many ways to analyze this circuit, depending one what techniques you have available to you. What circuit analysis techniques do you know so far?