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QUESTION: Hello Steve,
I was wondering if you could please help me with solving the current and voltage for these diagrams. I have attached an image for you below as well :)
i understand how to calculate the voltage of all the lamps in the first circuit, but I do not understand why the current is calculated the way it is said so in the answers.
I am also trying to calculate the voltage and current across the lamps in the second diagram but keep getting the voltage answer for the lamps E - J wrong as well as the currents for B,C, and D. May i please ask you kindly to help me on these two questions?
Thank you very much.

ANSWER: Hello Duffy,

5. Apparently we are to assume all 7 resistors are the same value. Call it R5.

Voltage: Resistor A has 8 V across it.
Resistors B&C each have 4 V across them.
Resistors D&E&F&G each have 2 V across them.

Current: If A has current I thru it, B&C have I/2 thru them, and D&E&F&G have I/4 thru them.
7 A = I + I/2 + I/4 = 1.75*I
I = 7 A/1.75 = 4 A
So B&C have 2 A thru them and D&E&F&G have 1 A thru them.

Out of curiosity, what is the value of R5? Look at the voltage across and the current thru resistor A.
R5 = 8 V / 4 A = 2 Ohms

6. Let these resistors be R6. I am puzzled by the "12 V" on the circuit diagram near resistor A. I don't know where the 0 V reference point is, so I will ignore that 12 V notation.

The equivalent resistance of the E-J combination is (2/3)*R6 = 0.67*R6.
The equivalent resistance of the B-D combination is
R6*2*R6/(R6+2*R6) = 2*R6/3 = 0.67*R6.
So then we have in series A, E-J, and B-D. The total circuit resistance is therefore 2.33*R6. It says that the current is 6 A, so
6 A = 28 V/(2.33*R6)
2.33*R6 = 28 V/6 A
Therefore R6 = 28/(2.33*6) = 2 Ohms

Voltage: So resistor A has 6 A*2 ohms = 12 V across it.

Look at the E&F combination. You would find 1/3 of the 6 A (2 A) thru that path, so the voltage across the pair E&F would be 2 A*2*2 Ohms = 8 V.
So E & F would each have 4 V across them.
Same is true of G&H and I&J.

The above has found 12 V across A and 8 V across the combination E-J, they total to 20 V. The battery is 28 V, so the remaining 8 V is dropped across the B-D combination.
B has 8 V across it and C & D each have 4 V across them.

Current: A has 6 A thru it.
Already said above is that the 6 A divides in thirds going thru the E-J combination. So each of those 6 resistors have 2 Amps going thru them.
Resistor B has 8 V across it, so the current thru it is 8 V/2 Ohms = 4 Amps.
The circuit current is 6 A, so C&D must have 6-4 = 2 Amps thru them.

To avoid this being longer than it is, I made some leaps. Not knowing how comfortable you are in these operations, I worry that some of my leaps may have left you puzzled. Feel free to ask followups.

I hope this helps,
Steve

---------- FOLLOW-UP ----------

QUESTION: Thank you very much Steve! This is helping me a lot! I know fully understand how you reached the answers for the voltages and current through number 5  but i am quite confused on the working out of the number 6 diagram as I am not very good with resistors and don't quite understand the first step you have taken out for the process in this specific diagram. If I may do so, can i kindly ask you to please explain the process of the number 6 diagram slightly more in depth as my knowledge on this topic isn't the best.
Thank you,
Duffy

Answer
Hello Duffy,

I have pasted in my original steps leading to the solutions for #6. I have numbered each of my steps. I then inserted support immediately under each step.

1. The equivalent resistance of the E-J combination is (2/3)*R6 = 0.67*R6.
E-J consists of 3 parallel paths (or branches), each path has 2 series resistors. Focus on resistors E & F. They are in series, and since series resistances just add (like the 2 zig-zag symbols combined so that it was 2X in length), the equivalent resistance of them is 2*R6. So each of the 3 parallel paths have equivalent resistance of 2*R6. (If you have 2 equal resistors in parallel, the equivalent resistance is 1/2 of the individual resistance. That is because when the electrons come to a fork in the road, they each have to choose which branch to take. If the 2 branches are equally easy to go thru, half will go left and the other half will go right. With 3 equal parallel paths, the same logic says that the equivalent resistance is 1/3 of the individual resistance. I showed above that the individual branches are each 2*R6, so the equivalent resistance of 3 of those in parallel would be (1/3)*2*R6. (A more rigorous way to calculate that is to take 2 of the 3 paths and find the equivalent resistance of them. Call that ReqTemp. Use the "product over sum" method explained in the next section. Then take the remaining branch and find the equivalent resistance of the 3rd path and ReqTemp. Again, use product over sum. That result is the equivalent of the E-J combination.)

2. The equivalent resistance of the B-D combination is
R6*2*R6/(R6+2*R6) = 2*R6/3 = 0.67*R6.
Here we have 2 parallel paths, one with resistance R6 and the other with resistance 2*R6. Here I used the "product over sum" method of calculating parallel resistance. You may not have seen that. It is valid when you are considering only 2 parallel paths. Assume we have R1 and R2 in parallel. The textbook method of calculating the equivalent resistance for any number in parallel, applied to these 2 resistors is:
1/Req = 1/R1 + 1/R2
Solving for Req by finding common denominator of right side, summing right side, and then inverting both sides.
1/Req = R2/(R1*R2) + R1/(R1*R2) = (R2+R1)/(R1*R2)
Req = R1*R2 / (R2+R1)
So Req ends up being the product of the 2 individual resistors over the sum of them. Notice that if the 2 are equal, it ends up being 1/2 of the individual value - supporting what I said
in my explanation above about how I got the equivalent resistance of the E-J combination.

3. So then we have in series A, the E-J, and B-D. The total circuit resistance is therefore 2.33*R6.
A is just R6, the equivalent resistance of combination E-J, and the equivalent resistance of B-D. These are in series, so the resistance of the entire network is the sum of
A which we are calling R6
combination E-J which I showed is 0.67*R6 and
combination B-D which I showed is also 0.67*R6
So the sum is 2.33*R6. That is the equivalent resistance of all 10 resistors wired up in that way.

4. It says that the current is 6 A, so
6 A = 28 V/(2.33*R6)
2.33*R6 = 28 V/6 A
Therefore R6 = 28/(2.33*6) = 2 Ohms
Taken one line at a time, 6 A = 28 V/(2.33*R6) comes from an application of Ohm's Law.
This is one of the forms of the Law: I = V/R
And then the other lines are just solving for R6. The problem didn't ask the value of the resistors, but I wanted it for the next step.

5. Voltage: So resistor A has 6 A*2 ohms = 12 V across it.
This is also an application of Ohm's Law.

6. Look at the E&F combination. You would find 1/3 of the 6 A (2 A) thru that path, so the voltage across the pair E&F would be 2 A*2*2 Ohms = 8 V.
So E & F would each have 4 V across them.
This step, "find 1/3 of the 6 A" is due to the principle I used and explained in step 1 above.
Same is true of G&H and I&J.
This step, "So E & F would each have 4 V" is due to symmetry (1/2 of the 8 V) or Ohm's Law (2 A*2 Ohms), whichever you see better.
So the combination E-J has 8 V across it.

7. The above has found 12 V across A and 8 V across the combination E-J, they total to 20 V. The battery is 28 V, so the remaining 8 V is dropped across the B-D combination.
B has 8 V across it and C & D each have 4 V across them.
This step, "so the remaining 8 V" is an application of Thevenin's Rule. In study of voltages increases and decreases as you look at a loop in a circuit, the increases must equal the decreases. A battery (circulating minus to plus) is an increase and the voltage found across resistors is a decrease. This is a quick and dirty statement of Thevenin's Rule. You may not have studied this in any detail yet. I hope the quick and dirty statement is clear and seems justified.

8. Current: A has 6 A thru it.
This is stated on the schematic.

9. Already said above is that the 6 A divides in thirds going thru the E-J combination. So each of those 6 resistors have 2 Amps going thru them.
This was supported this above.

10. Resistor B has 8 V across it, so the current thru it is 8 V/2 Ohms = 4 Amps.
Ohm's Law

11. The circuit current is 6 A, and 4 Amps is the current thru resistor B, so C&D must have 6-4 = 2 Amps thru them.
The total current of 6 A divides into 2 branches at this point. What doesn't go thru B must be going thru the other branch.

I hope this helps,
Steve

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Steve Johnson

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I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.

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I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

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BS Physics, North Dakota State University
MS Electrical Engineering, North Dakota State University

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