electrical and electronic question

Thread Starter

1993terry

Joined Oct 26, 2018
8
ive been struggling on this question for a long time now i believe its KCL not 100% can anyone explain how to get to the answer

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ericgibbs

Joined Jan 29, 2010
18,767
hi,
OK 1500R , but the 1/Rt is not the Equation that is expected.??
So now you know R2||R3, what is the total Series resistance across the 12v battery.?
E
 

WBahn

Joined Mar 31, 2012
29,979
so 1/2000+1/6000=1/Rt =1500 ?
You need to get into the habit of properly tracking your units. Treat any dimensioned quantity as the product of a magnitude and a unit and then just apply the normal rules of arithmetic and algebra accordingly.

Doing so will let you detect most errors you make almost immediately.

So in this case, being more explicit than is needed to illustrate the point, we would have

1/Rt = 1/R2 + 1/R3

1/Rt = R3/(R2·R3) + R2/(R2·R3)

1/Rt = (R2 + R3) / (R2·R3)

Rt = (R2·R3) / (R2 + R3)

Notice that, up till now, everything has been done symbolically (i.e., "no numbers"). But the symbols carry units -- in this case Rt, R1, R2, and R3 all have units of resistance. At each step we can check that the units work out. For instance, in the next-to-last line we have 1/resistance on the left and on the right we have resistance divided by resistance-squared, which yields 1/resistance. So that works out. Similarly, the last line works out to units of resistance on both sides.

Now we plug in the values -- still tracking the units.

Rt = ( (2 kΩ)·(6 kΩ) ) / ( 2 kΩ + 6 kΩ )

Rt = ( 12 kΩ² ) / ( 8 kΩ )

Rt = 1.5 kΩ

A very common error that people make happens when working with the parallel resistance formula (or other formulas with similar form) and they get the numerator and denominator swapped, often because they are getting sloppy and doing too much in their head. So they might write down

Rt = (2 + 6) / (2 * 6) = 0.667 kΩ

What they are doing is ignoring units and then tacking the units that they WANT the answer to have onto the end. In doing so, they miss out on a golden opportunity to catch their mistake at the point it is made.

Rt = ( 2 kΩ + 6 kΩ ) / ( (2 kΩ)·(6 kΩ) )

Rt = ( 8 kΩ ) / ( 12 kΩ² )

Rt = 0.667 kΩ^-1

The units don't work out, so they would then KNOW they had made a mistake.

Properly tracking units is perhaps the single most powerful error detection and correction tool available to the engineer. Learn to use it and you will see your grades go up significantly and avoid lots of silly mistakes (which we ALL make) going uncaught until your supervisor or customer finds them.
 

Thread Starter

1993terry

Joined Oct 26, 2018
8
What current does that give for R1 and Rt?
V/R1 = 12/5000 =2.4x10^-3
V/Rt = 12/1500=8x10^-3

i was confused if they could be now added in series and have 5000+1500-6500R and then have 12/6500=1.84x10^-3

sorry for long reply can only post 5 an hour
 

WBahn

Joined Mar 31, 2012
29,979
V/R1 = 12/5000 =2.4x10^-3
V/Rt = 12/1500=8x10^-3
You are making one of the classic mistakes and just throwing whatever V and R you happen to find at Ohm's Law and hoping that the result means something.

Ohm's Law relates the resistance of a resistor to the voltage across THAT resistor and the current through THAT resistor.

The 12 V is NOT the voltage across ANY of those resistors (unless one of them happens to be either 0 Ω).

i was confused if they could be now added in series and have 5000+1500-6500R and then have 12/6500=1.84x10^-3
I have no idea what "5000+1500-6500R" is.

But, yes, once you have the parallel combination of R2 and R3 (let's call it R23 and reserve Rt for the total circuit resistance), then you can add R1 and R23 to get Rt.

sorry for long reply can only post 5 an hour
That's an anti-spammer limitation on New Members that will go away once you get 10 posts.
 

Thread Starter

1993terry

Joined Oct 26, 2018
8
You are making one of the classic mistakes and just throwing whatever V and R you happen to find at Ohm's Law and hoping that the result means something.

Ohm's Law relates the resistance of a resistor to the voltage across THAT resistor and the current through THAT resistor.

The 12 V is NOT the voltage across ANY of those resistors (unless one of them happens to be either 0 Ω).



I have no idea what "5000+1500-6500R" is.

But, yes, once you have the parallel combination of R2 and R3 (let's call it R23 and reserve Rt for the total circuit resistance), then you can add R1 and R23 to get Rt.



That's an anti-spammer limitation on New Members that will go away once you get 10 posts.
when you say add R1 and R23 together, thats what i was meaning by adding the 1500(R23) and (R1)5000 resistor together to get 6500, ?
is that different to what you are talking about ?
 

WBahn

Joined Mar 31, 2012
29,979
when you say add R1 and R23 together, thats what i was meaning by adding the 1500(R23) and (R1)5000 resistor together to get 6500, ?
is that different to what you are talking about ?
That's what I was referring to, it just didn't jive with what you wrote, except R23 is NOT 1500, it is 1500 Ω. Start using units properly -- it could literally save your life some day.

Another comment I should have made your prior post is that another good practice to always ask if your answers make sense.

Your results claim that the current in R1 is 2.4 mA but that the combined currents in R2 and R3 total to 8 mA, while the current being delivered by the voltage supply is 1.84 mA.

Do those results make sense?

If not, then it's time to step back and not go forward until they do. You've likely make a mistake somewhere. Perhaps it was conceptual (like not properly applying Ohm's Law), or perhaps it was just a silly math error. But it's a big red warning flag telling you to stop and review your work because going forward is most likely only going to be a bunch of wasted effort.
 

Thread Starter

1993terry

Joined Oct 26, 2018
8
That's what I was referring to, it just didn't jive with what you wrote, except R23 is NOT 1500, it is 1500 Ω. Start using units properly -- it could literally save your life some day.

Another comment I should have made your prior post is that another good practice to always ask if your answers make sense.

Your results claim that the current in R1 is 2.4 mA but that the combined currents in R2 and R3 total to 8 mA, while the current being delivered by the voltage supply is 1.84 mA.

Do those results make sense?

If not, then it's time to step back and not go forward until they do. You've likely make a mistake somewhere. Perhaps it was conceptual (like not properly applying Ohm's Law), or perhaps it was just a silly math error. But it's a big red warning flag telling you to stop and review your work because going forward is most likely only going to be a bunch of wasted effort.
yes i see what your saying ,

so if i do 12v/6500Ω i get 1.84mA so that should be the total resistance,
if i do 12v/1500Ω i get 8mA so that means that R1 should equal 1.04mA ?
but i would say using ohms law V/R=I and would do 12v/5000Ω to get 2.4mA which like your saying cant be correct im not sure where ive went wrong am i doing the correct calculations and my math is wrong or is the maths correct and using ohms law incorrect ?
 

WBahn

Joined Mar 31, 2012
29,979
yes i see what your saying ,

so if i do 12v/6500Ω i get 1.84mA so that should be the total resistance,
if i do 12v/1500Ω i get 8mA so that means that R1 should equal 1.04mA ?
but i would say using ohms law V/R=I and would do 12v/5000Ω to get 2.4mA which like your saying cant be correct im not sure where ive went wrong am i doing the correct calculations and my math is wrong or is the maths correct and using ohms law incorrect ?
You want to use V/R=I and you are just grabbing the nearest V, which happens to be 12 V, without any regard as to whether that 12 V is actually the voltage appearing ACROSS the 5000 Ω resistor.

But it's NOT the voltage ACROSS (i.e., the voltage difference from one side of that resistor to the other side of that resistor).

The 12 V is simply the voltage across the voltage supply.

Let's throw some node labels at this thing as well as some voltage and current declarations, so that we can talk very specifically about this circuit.

Figure1.PNG

The green labels are the node labels.

If the voltage at Node A is Va and the voltage at Node B is Vb, then the voltage at Node A relative to the voltage at Node B is Vab. In other words, when using subscripts to identify nodes, we have

Vxy = Vx - Vy

The numbered voltage declarations may be written as voltage differences between the nodes in which it is the voltage at the '+' node relative to the voltage at the '-' node.

V1 = Va - Vb = Vab

What are the others?

What does Ohm's Law say about the relationships between the numbered currents and the numbered voltages?

What does KCL say about the relationships among the currents?

What does KVL say about the relationships among the voltages?

This will allow you to develop a set of equations that will let you solve for the voltages and currents without reducing the circuit to a single equivalent resistance and then walking the solution back out.

I can't tell for sure, but that seems to be about the level that this problem is trying to take you. Perhaps I'm a bit too early for where you are right now.
 
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