Hey, all. The attached PDF shows the questions I had on my quiz earlier today.

On #1, all resistors are 3.3k.
#2, Current at R1 = 2mA at R2 = 1.14mA at R3 = 0.42mA


The following are my answers:

1A. Current through R2 = 5.05mA
1B. Total current is 10.1mA

2A.
2B. 2.58mA @ R4

3. RT = 2.02kΩ

4A. Total current 939mA
4B. Current @ R1 = 273mA
Current @ R2 = 333mA
Current @ R3 = 333mA

5. Current @ R1 = 1.001mA
Current @ R2 = 2.001mA



I didn't have enough time to get the values of the resistors on question #2.


BTW, that's my professor's handwriting...


Thanks in advance.

 

On #1, all resistors are 3.3k.
#2, Current at R1 = 2mA at R2 = 1.14mA at R3 = 0.42mA

Wrong.

First, you should add up all the parallel resistors (and they are all parallel). So \(\frac{1}{3.3k}+\frac{1}{3.3k}+\frac{1k}{3.3k}=1.1k\)

Now to find the total current, \(\frac{50V}{1.1k}=45.5mA\)

For each individual resistor, \(\frac{50V}{3.3k}=15.2mA\).

So add them up just to make sure the total current is correct: \(15.2mA+15.2mA+15.2mA=45.6mA\). That is indeed correct. So each resistor has 15.2mA of current flow and the total current is about 45.5mA.

 

So, all three resistors are in parallel? I don't see that.

 

Scratch that, they are.

Look at it this way, they share a common node at each end

 

So, all three resistors are in parallel? I don't see that.

Yes indeed, they are. Take a look at R1 and R2, just move R1 so it's completely vertical and you have both resistors in parallel. Furthermore, move R3 so it's vertical too and you'll see that all three resistors are in parallel. Understand?

 

Let me take another look...

 

For your second quiz question, what current values are listed for R2 and R3? It's difficult to read them, so could you list them on a post?

Austin

 

I have just noticed how they are all in parallel....

This is great...

And the currents are listed on my first post.

I total = 5mA
I @ R1 = 2mA
I @ R2 = 1.14mA
I @ R3 = 0.42mA

 

And the currents are listed on my first post.

I total = 5mA
I @ R1 = 2mA
I @ R2 = 1.14mA
I @ R3 = 0.42mA

Now that you agree the resistors are connected parallel, don't you agree your original calculations might need revising?

 

Now that you agree the resistors are connected parallel, don't you agree your original calculations might need revising?

That was to ELECTRONERD in reference to the second question on the quiz.

I have a feeling I bombed this quiz.

Thanks for the help guys.

 

I have just noticed how they are all in parallel....

This is great...

And the currents are listed on my first post.

I total = 5mA
I @ R1 = 2mA
I @ R2 = 1.14mA
I @ R3 = 0.42mA

Ok, I think you made a mistake listing the total current. You say the total current is 5mA but in the PDF it has 50mA total going into the circuit. I'm assuming it is 5mA based on each individual current values at each resistor.

First you add them up, \(2mA + 1.14mA + 0.42mA = 3.56mA\) So obviously you would subtract the current through each resistor which we summed up to be 3.56mA from the total current to find the current value for R4. \(5mA - 3.56mA = 1.44mA\) Therefore we can assume that the current through R4 is 1.44mA and that all the currents through each resistor sum up to the total current (5mA). \(2mA + 1.14mA + 0.42mA + 1.44mA = 5mA\) That is true and the current through R4 is indeed 1.44mA.

2A.
2B. 2.58mA @ R4

Do you see how this is wrong? Also, I'm unsure what they are asking for for 2a. Do they actually want the value of R4 in ohms? If they do, it's interesting that they didn't have you find the current through R4 before finding the actual resistor value.