Im trying to solve this one, but without any luck. My approach is as follows:
I added my own currentflow, called Ix. Then I would just simply solve:
Ix - I2 - I1 = 0 (KCL)
5V - 4700*Ix - 2V - 1200*I2 = 0V (KVL)
2V - 1000*I1 + 1200*I2 (KVL)

Is this the correct approach? After solving this i get I2 = 1.357 mA and I1 =-0.525mA, while the answer model says 1.33mA and -0.555mA...

 

Hi kaz,
As this homework, we cannot give an answer, only guidelines.
But the model answer is correct.
Please show all your calculations.
E

 

Below is the LTspice sim of the circuit, verifying the answer model values:

 

@kazkaz welcome! i'm also new here. after checking your KCL and KVL. it is correct. I think you only made a calculation error. I can find the same calculation as @crutschow 's LTspice value.

 

View attachment 302285
Im trying to solve this one, but without any luck. My approach is as follows:
I added my own currentflow, called Ix. Then I would just simply solve:
Ix - I2 - I1 = 0 (KCL)
5V - 4700*Ix - 2V - 1200*I2 = 0V (KVL)
2V - 1000*I1 + 1200*I2 (KVL)

Is this the correct approach? After solving this i get I2 = 1.357 mA and I1 =-0.525mA, while the answer model says 1.33mA and -0.555mA...

First off, your answers are close to the the provided answers. 1.357 mA is about 2% too large and -0.525 mA is off by about 5.4%. So the problem could be with nothing more than how you carried out your work and whether you carried enough significant digits in your work to get an answer that is good to three sig figs. If you are going to work with numerical values at intermediate steps, then you should carry at least two additional sig figs than what you want the answer to be good for. You should be able to be in agreement with the exact answer well within 1%.

I'm going to tweak your equation to reflect the proper use of units, but other than that I'm not changing them except to turn the third one into an equation instead of just an expression.

Ix - I2 - I1 = 0 (KCL)
5 V - (4700 Ω)*Ix - 2 V - (1200 Ω)*I2 = 0 V (KVL)
2 V - (1000 Ω)*I1 + (1200 Ω)*I2 = 0 V (KVL)

The first thing to do is to see if the answers you got actually satisfy the equations you started with. Even if the equations are wrong, your answers should satisfy them.

(2 V) - (1 kΩ)(-0.525 mA) + (1.2 kΩ)(1.357 mA) = 3.10 V

Notice that this is no where near 0 V, so something is massively wrong.

Are you sure you don't have I1 and I2 swapped in your answers and that you meant that you got I1 = 1.357 mA and I2 =-0.525mA,

If so, that makes it

(2 V) - (1 kΩ)(1.357 mA) + (1.2 kΩ)(-0.525 mA) = 13 mV

That's in the round-off error realm.

Ix = I1 + I2 = 1.357 mA + -0.525 mA = 832 mA

(5 V) - (4.7 kΩ)(832 mA) - (2 V) - (1.2 kΩ)(-0.525 mA) = -280 mV

This one, while in the ballpark, is quite a bit more off than we should accept.

My guess is that you just either didn't carry enough sig figs through your work or you made a silly math error (we all do). Since you didn't show the details of your work, I can't guess any further than that. But I want to at least thank you for showing your setup equations and annotating your drawing to define the new variable, Ix, that you introduced.

Let's see if their answers come closer:

(2 V) - (1 kΩ)(1.33 mA) + (1.2 kΩ)(-0.555 mA) = 4 mV
Ix = I1 + I2 = 1.33 mA + -0.555 mA = 775 mA
(5 V) - (4.7 kΩ)(775 mA) - (2 V) - (1.2 kΩ)(-0.555 mA) = -22.5 mV

That's certainly much, much better, but I still don't like it. We have voltages on the order of 1 V to 5 V and we really should be able to get "closure" in our results of 0.1% (which is what three sig figs is basically claiming), so we should be looking for errors of 5 mV or less.

The first obvious likely culprit is that 1.33 mA. I'm willing to bet that it should be 1.333 mA. A common rule is that a leading 1 in a value doesn't count as a significant figure (some people say that neither a leading 1 nor a leading 2 should be counted).

Instead of going through and solving the system of equations myself, I'm going to check the results that the simulation produced above.

I1 = 1.334 mA
I2 = 0.555 mA

Notice that the only difference is that we have reported I1 to one more sig fig because we aren't counting the leading 1.

(2 V) - (1 kΩ)(1.334 mA) + (1.2 kΩ)(-0.555 mA) = 0 mV
Ix = I1 + I2 = 1.33 mA + -0.555 mA = 779 mA
(5 V) - (4.7 kΩ)(779 mA) - (2 V) - (1.2 kΩ)(-0.555 mA) = 4.7 mV

Now, by properly carrying three sig figs in all of our results, our closures are within 5 mV, which is consistent with 0.1% accuracy on values in the 5 V range.

 

View attachment 302285
Im trying to solve this one, but without any luck. My approach is as follows:
I added my own currentflow, called Ix. Then I would just simply solve:
Ix - I2 - I1 = 0 (KCL)
5V - 4700*Ix - 2V - 1200*I2 = 0V (KVL)
2V - 1000*I1 + 1200*I2 (KVL)

Is this the correct approach? After solving this i get I2 = 1.357 mA and I1 =-0.525mA, while the answer model says 1.33mA and -0.555mA...

Hello,

Sometimes the text of the problem is not correct so it pays to verify that.
I went as far as to verify I1 and got 1.334488734835ma which can be rounded to 1.33ma as your text shows so it is correct for that current.
I did not check I2 yet.
The approach I used was not nodal however even though I usually use that method. I used superposition combined with the properties of current division between two resistors in parallel. It's good to have a second method to check your first method with these problems. If you get a different result with one method than the other you must have made a mistake somewhere. This idea reduces your reliance on other people to check your work and almost guarantees a good mark on every test where you don't get to ask anyone else to verify your results.

 

It's good to have a second method to check your first method with these problems. If you get a different result with one method than the other you must have made a mistake somewhere. This idea reduces your reliance on other people to check your work and almost guarantees a good mark on every test where you don't get to ask anyone else to verify your results.

I think a better approach is to verify the results directly from the problem itself, which is almost always possible even in real-world engineering situations. This usually takes considerably less effort and time than solving the problem from scratch using a different method, a consideration that is of particular importance on most exams.

In this case, I would probably approach it along the following lines.

Let's first consider the answers as originally offered by the TS: " i get I2 = 1.357 mA and I1 =-0.525mA, "
Call the top-right node Vo.

Given the answers, we can calculate Vo two different ways:

Vo = (1.2 kΩ)(I2) + 2 V = (1.2 kΩ)(1.357 mA) + 2 V = 3.628 V

Vo = (1 kΩ)(I1) = (1 kΩ)(-0.555 mA) = -0.555 V

BIG RED FLAG!

At this point, hopefully it wouldn't take long to realize that we swapped I1 and I2 in reporting the answer. So now we do it again:

Vo = (1.2 kΩ)(I2) + 2 V = (1.2 kΩ)(-0.555 mA) + 2 V = 1.334 V

Vo = (1 kΩ)(I1) = (1 kΩ)(1.357 mA) = 1.357 V

These don't agree as close as they should, but they are close, possibly indicating some sloppy match. Worth going back and seeing if we can spot the mistake and fix it, or at least carry out our computations with an extra couple of sig figs. Doing that, we would get a revised value for I1 of 1.334 mA (I2 wouldn't change) and so now our two paths to Vo agree.

Finally, we can close the loop by either using Vo to calculate Ix, or using Ix to get yet another value for Vo along the remaining path. Let's do that:

Vo = 5 V - (Ix)(4.7 kΩ) = 5 V - (I1 + I2)(4.7 kΩ) = 5 V - (1.334 mA + -0.555 mA)(4.7 kΩ) = 1.339 V

This is consistent with our goal of three sig figs.

If we didn't revise our I1 before carrying out this final check, we would have gotten:

Vo = 5 V - (Ix)(4.7 kΩ) = 5 V - (I1 + I2)(4.7 kΩ) = 5 V - (1.357 mA + -0.555 mA)(4.7 kΩ) = 1.231 V

This is sufficiently not in agreement with the other two values for Vo, 1.334 V and 1.357 V, that I would want to explore it further.

On the other hand, these are close enough that, on something like an exam, I might let it be and move on to other problems and come back to this one if time permits. Hopefully I've done enough to claim at least most of the points.

 

I think a better approach is to verify the results directly from the problem itself, which is almost always possible even in real-world engineering situations. This usually takes considerably less effort and time than solving the problem from scratch using a different method, a consideration that is of particular importance on most exams.

In this case, I would probably approach it along the following lines.

Let's first consider the answers as originally offered by the TS: " i get I2 = 1.357 mA and I1 =-0.525mA, "
Call the top-right node Vo.

Given the answers, we can calculate Vo two different ways:

Vo = (1.2 kΩ)(I2) + 2 V = (1.2 kΩ)(1.357 mA) + 2 V = 3.628 V

Vo = (1 kΩ)(I1) = (1 kΩ)(-0.555 mA) = -0.555 V

BIG RED FLAG!

At this point, hopefully it wouldn't take long to realize that we swapped I1 and I2 in reporting the answer. So now we do it again:

Vo = (1.2 kΩ)(I2) + 2 V = (1.2 kΩ)(-0.555 mA) + 2 V = 1.334 V

Vo = (1 kΩ)(I1) = (1 kΩ)(1.357 mA) = 1.357 V

These don't agree as close as they should, but they are close, possibly indicating some sloppy match. Worth going back and seeing if we can spot the mistake and fix it, or at least carry out our computations with an extra couple of sig figs. Doing that, we would get a revised value for I1 of 1.334 mA (I2 wouldn't change) and so now our two paths to Vo agree.

Finally, we can close the loop by either using Vo to calculate Ix, or using Ix to get yet another value for Vo along the remaining path. Let's do that:

Vo = 5 V - (Ix)(4.7 kΩ) = 5 V - (I1 + I2)(4.7 kΩ) = 5 V - (1.334 mA + -0.555 mA)(4.7 kΩ) = 1.339 V

This is consistent with our goal of three sig figs.

If we didn't revise our I1 before carrying out this final check, we would have gotten:

Vo = 5 V - (Ix)(4.7 kΩ) = 5 V - (I1 + I2)(4.7 kΩ) = 5 V - (1.357 mA + -0.555 mA)(4.7 kΩ) = 1.231 V

This is sufficiently not in agreement with the other two values for Vo, 1.334 V and 1.357 V, that I would want to explore it further.

On the other hand, these are close enough that, on something like an exam, I might let it be and move on to other problems and come back to this one if time permits. Hopefully I've done enough to claim at least most of the points.

Hi,

Maybe that is almost the same as another way I like to do this (but maybe not).
With nodal, the first method, we get all the node voltages. Since we have all the node voltages the individual element currents are easy to calculate. We can then just sum the currents into each node and make sure that they add up algebraically to zero.
When I do checks like this I like to use at least 8 digits, but I realize not everyone likes to be that careful.

The method I used was easy for me it only takes a few different concepts to get the answer to I1:
1. Know how to calculate the total resistance of two parallel resistors.
2. Know how to use superposition with two voltage sources.
3. Know how to calculate the current in each resistor of a two-resistor parallel circuit knowing the total current through them both (current division).
I saw that solution as soon as I looked at the circuit so I went with that just for a quick check.