How can I determine the current in all resistors in a circuit with given values?

In summary, the goal of this problem was to find the current in all resistors given the current in one resistor and using various laws and principles including the parallel and series resistor laws, KCL, and KVL. The final solution involved calculating the voltage across the 13.8 ohm resistor and using it to find the current in the parallel combination of the 8.45 and 4.11 ohm resistors. This current, along with the current in the 13.8 ohm resistor, was then used to find the total current passing through the 12.5 and 15 ohm resistors. Finally, Ohm's Law was used to calculate the voltage drops across these resistors.
  • #1
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Homework Statement



Wldlu.jpg

The current in the 13.8 resistor = .780 A. Goal is to find the current in all resistors.


Homework Equations



Parallel resistor law - R total = 1 / (1/R1 + 1/R2 + ...)
Series resistor law - R total = R1 + R2 + ...
KCL - sum of currents flowing into a node must be = to sum flowing out
KVL = directed sum of voltages in a closed circuit = 0

The Attempt at a Solution



Okay, quick preface... a friend randomly asked me if I had any idea how to do this. I haven't been to school in about 6 years, my only knowledge on the subject was what I taught myself way back then and I'm just interested in how this works for personal understanding as I honestly can't recall what's what and what I'm doing wrong and the writeups in Wiki/random tutorial sites are not helping me understand...

First, given the 13.8 resistor with .780 A, I figure out the voltage is 10.764. Applying this to the parallel resistor gives me .626 A. Using the parallel resistor law, I simplify the two to come up with a 7.66 resistor which I name R3. The two currents combined = 1.406 A or I3.

Now I add the other 15 and 12.5 resistors with the serial law for 27.5 or R1. I do the same for 8.45 and 4.11 for 12.56 or R3.
From here, I'm lost...
I call the second half of my current circuit Loop 2 and using KVL, I *think* R3I3 - R2I2 = 0. Going through this formula gives me a current of .858 for I2. Now, with KCL... I2+I3 = I1 = 2.264.

Now when applying KVL to loop 1, R1I1 - R3I3 should = 0, but that's not what I get.
I'm almost positive I'm messing something up when it comes to KVL but I can't tell what and all the examples I've found online haven't helped...
 
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  • #2
I didn't check your numbers, but your method sounds correct until your final application of KVL. It's not working because you've ignored the voltage across the battery. You also need to be careful with the signs. The terms for both R1 and R3 should have the same sign based on how you calculated the currents.
 
  • #3
Firstly, how do the 8.45- and 4.11-ohm resistors relate, positionally, to the 13.8- and 17.2-ohm resistors?
 
  • #4
The two series resistors on the right are in parallel with the 13.8 ohm resistor.

So, you can work out the current in this combination since they have the same voltage across them as the 13.8 ohm resistor.

This gives you 3 currents to add up to the total current which has to pass through the 12.5 ohm and 15 ohms resistors causing voltage drops. So these voltage drops add up, with the original 10.764 volts, to give the battery voltage, (although this isn't asked for).

This probably isn't a KVL problem. It is just Ohm's Law.
 
  • #5
There you go. The problem turns out to be quite a bit easier than it first looks.
 

1. What is resistance in circuits?

Resistance in circuits is the opposition to the flow of electrical current. It is measured in ohms and is caused by factors such as the material of the wire and its length and thickness.

2. How is resistance calculated?

Resistance is calculated using Ohm's Law, which states that resistance is equal to the voltage divided by the current. It can also be calculated using the formula R = ρ * (L/A), where ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area.

3. How does resistance affect the flow of current?

Resistance affects the flow of current by reducing it. The higher the resistance, the lower the current flow. This is because resistance causes a voltage drop, which decreases the potential difference and therefore reduces the flow of current.

4. What factors affect resistance in a circuit?

The main factors that affect resistance in a circuit are the material of the wire, its length and thickness, and the temperature. Materials with a higher resistivity, longer wires, and thinner wires will have higher resistance. Additionally, as the temperature increases, the resistance of most materials also increases.

5. How can resistance be reduced in a circuit?

Resistance can be reduced in a circuit by using materials with lower resistivity, shorter and thicker wires, and keeping the temperature low. Additionally, using conductors with a larger cross-sectional area, such as copper, can also help reduce resistance in a circuit.

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