Given this diagram, I am asked to calculate the current at resistor R2. Can I just calculate it using Ohm's law (I=20V/3Ω) or is the current affected by resistor R1?

The current through R2 is depending on all the resistors! You will have to solve this stepwise. First calculate the total current in the system (20V over the total resistance), then you can calculate how much current goes through the branch R4+R3 and how much through R2.

To understand which resistors have the same current always move from your sources "+" to its "-" along the wires.
In this case all the current there is goes through R1, as you need to go through it to get from + to - Then you can go either through R3 or through R4 and R2. This means that R3 has a different current than R4 and R2. The sum off currents going through R3 and R2,R4 must be the same as the one going through R1 (so I\_1 = I\_3+I\_2,4), which is one of Kirchhoffs Laws.

if you look closely, this is a voltage divider. sum r3 and r4, remove r2 and redraw the circuit. insert r2, can you see it as the 'load'? then you could just apply thevenin and get the current for an arbitrary load.
btw are you taking Circuits I?

You can, but only if R1 does not exist.
[https://i.imgur.com/lM0G1SN.jpg](https://i.imgur.com/lM0G1SN.jpg)
So you need to figure out what is the **voltage across the parallel resistors section**. Then you can apply that equation.
The arrow colors indicate the different currents that go through them

There are some concepts that you don't seem to have been taught yet. Kirchhoffs current law states that the “total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node“. So the current leaving R1 is divided into two branches. One branch goes through R2 and one goes through R3 and R4. The way to figure out how much current is in the total circuit is to understand how much total resistance there is in the total circuit.
Resistors in series add normally, resistors in parallel add by conductance. [https://en.wikipedia.org/wiki/Siemens\_(unit)](https://en.wikipedia.org/wiki/Siemens_(unit)) This is the inverse of resistance.
So the first thing I would do is add R3 and R4 together which gets us 6Ω. Then we add the parallel resistors R2 + (R3+4). See this link for a more thorough explanation of adding resistors in parallel. [https://www.electronics-tutorials.ws/resistor/res\_4.html](https://www.electronics-tutorials.ws/resistor/res_4.html)
After calculating I come up with 2Ω of resistance for R2+3+4.
R1 + 2Ω gives a total circuit resistance of 10Ω. Now using ohms law, 20V/10Ω=2A.
Since amperage is the same in a series circuit and we've added the resistors to imagine them as a series circuit, we know that there is 2 amps of current going through our total circuit. There is 2 amps going though R1 and then then amperage is divided to go through the two branches. We know this because of Kirchoffs current law. So to get to the next step of finding out how much current is going through each branch, we will consider the voltage drops. 8Ω\*2A=16V and 2Ω\*2A=4V.
16 of the 20 total volts is being resisted by R1. The remainder of the circuit is only seeing 4 of the 20 volts. Now we know that the parallel branches are each seeing 4 volts.
Now that we have 4 volts and a 3 ohm resistor, we can divide **4V/3Ω=1.33A**. To double check we look at the other branch which also has 4 volts but has 2Ω+4Ω which is 6Ω. 4V/6Ω=.67A. If we add 1.33A+.67A=2A.

I feel like these exercises are completely unrealistic.
I knew nothing about this industry a year ago and now I'm building completely functional circuits for my own purposes without any education that I have never had to do any of these problems.
College is a scam.

The current through R2 is depending on all the resistors! You will have to solve this stepwise. First calculate the total current in the system (20V over the total resistance), then you can calculate how much current goes through the branch R4+R3 and how much through R2.

Thanks. Do R1 and R2 share a current? Do R3 and R4 have a seperate current?

To understand which resistors have the same current always move from your sources "+" to its "-" along the wires. In this case all the current there is goes through R1, as you need to go through it to get from + to - Then you can go either through R3 or through R4 and R2. This means that R3 has a different current than R4 and R2. The sum off currents going through R3 and R2,R4 must be the same as the one going through R1 (so I\_1 = I\_3+I\_2,4), which is one of Kirchhoffs Laws.

if you look closely, this is a voltage divider. sum r3 and r4, remove r2 and redraw the circuit. insert r2, can you see it as the 'load'? then you could just apply thevenin and get the current for an arbitrary load. btw are you taking Circuits I?

I'm in middle school, we just started learning circuits

um, now I don't know if my answer was appropriate 😅 did you see Thevenin's Theorem in class?

Nah, we really just started learning two weeks ago. I got an answer now, so thanks!

You can, but only if R1 does not exist. [https://i.imgur.com/lM0G1SN.jpg](https://i.imgur.com/lM0G1SN.jpg) So you need to figure out what is the **voltage across the parallel resistors section**. Then you can apply that equation. The arrow colors indicate the different currents that go through them

1.33 amps

There are some concepts that you don't seem to have been taught yet. Kirchhoffs current law states that the “total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node“. So the current leaving R1 is divided into two branches. One branch goes through R2 and one goes through R3 and R4. The way to figure out how much current is in the total circuit is to understand how much total resistance there is in the total circuit. Resistors in series add normally, resistors in parallel add by conductance. [https://en.wikipedia.org/wiki/Siemens\_(unit)](https://en.wikipedia.org/wiki/Siemens_(unit)) This is the inverse of resistance. So the first thing I would do is add R3 and R4 together which gets us 6Ω. Then we add the parallel resistors R2 + (R3+4). See this link for a more thorough explanation of adding resistors in parallel. [https://www.electronics-tutorials.ws/resistor/res\_4.html](https://www.electronics-tutorials.ws/resistor/res_4.html) After calculating I come up with 2Ω of resistance for R2+3+4. R1 + 2Ω gives a total circuit resistance of 10Ω. Now using ohms law, 20V/10Ω=2A. Since amperage is the same in a series circuit and we've added the resistors to imagine them as a series circuit, we know that there is 2 amps of current going through our total circuit. There is 2 amps going though R1 and then then amperage is divided to go through the two branches. We know this because of Kirchoffs current law. So to get to the next step of finding out how much current is going through each branch, we will consider the voltage drops. 8Ω\*2A=16V and 2Ω\*2A=4V. 16 of the 20 total volts is being resisted by R1. The remainder of the circuit is only seeing 4 of the 20 volts. Now we know that the parallel branches are each seeing 4 volts. Now that we have 4 volts and a 3 ohm resistor, we can divide **4V/3Ω=1.33A**. To double check we look at the other branch which also has 4 volts but has 2Ω+4Ω which is 6Ω. 4V/6Ω=.67A. If we add 1.33A+.67A=2A.

I feel like these exercises are completely unrealistic. I knew nothing about this industry a year ago and now I'm building completely functional circuits for my own purposes without any education that I have never had to do any of these problems. College is a scam.