Help, my teacher is dropping truth bombs about circuits at a

I apologize for the sideways picture, Iphone is cucking everyone out there.

Now, because R1 is in a series, I can just do I=E/R and get the current, which is also the total current because it's series, correct?

>Finding total current through circuit and total resistance

Find the combined resistance of the 3 parallel resistors, and then add the result to R1 to get the total resistance in the circuit. From there you can find the total current through the circuit using the equation V = IR.

>Finding current through each of the 3 parallel resistors
Find the PD across R1 (current through R1 is also total current through circuit like you said), and you can then find the collective PD across the 3 parallel resistors using the voltage divider rule.

You can then calculate the individual currents through each of the resistors using the current divider rule.

Now you have all the values you need to just solve for power etc

>>190073
I'm getting 148 Ohms for Rt, can you confirm? What is PD? So total current is found.

To use the voltage divider rule, would I use R1 and the Req of R2+R3+R4? Assuming This is what I do, I can label R2, R3, and R4 with the same voltage and then Do I= E/R, right?

>>190077
For some reason, when I do the reciprocal of the reciprocal formula, I get a different answer, so the new Rt from using the x-1 button on my calculator gave me 92 Ohms.

>>190077

My calculation:
Rt = R1 + [(1/80) + (1/60) + (1/80)]^-1
= 68 + (1/24)^-1)
= 68 + 24
= 92 Ohms

PD is potential difference, aka voltage.

Yes, you use the combined resistance of the 3 parallel resistors to find their combined voltage. However you cannot assume the voltages for R2, R3 and R4 are the same. First find the currents that go to R2, R3 and R4 using the current divider rule. Then you can find the voltage across each of these resistors using V = IR

>>190086
Wait sorry I don't think you need to use the voltage divider. Just minus the voltage across R1, ie. 46 - V1.

>>190089
Can you explain this a bit more, please? I thought that V1 was 46V since it's the first resistor that it goes through, but I'm confused as to what the voltage drop was for that, I also thought that all parallel resistors had the same voltage, but the current is what changes.

>>190057
R_net=R1 + R1R2R3/(R1R2+R1R3+R2R3)
I_total/R1 = V/[R1 + R1R2R3/(R1R2+R1R3+R2R3)]
V_R1 = I_total * R1
V_R2/R3/R4 = V - V_R1
I_R2 = V_R2 / R2
I_R3 = V_R3 / R3
I_R4 = V_R4 / R4
P_R1 = I_total^2 R1
P_R2 = (V - V_R1)^2 / R2
P_R3 = (V - V_R1)^2 / R3
P_R4 = (V - V_R1)^2 / R4

>>190096
I think I found what I did wrong, I used Vs/R1 to find Current total, which I don't think it's right because you can't use a total with something that isn't a total to find another value. So, with this corrected, the Current total is .5 A, now with this, i can find V1. Gonna calculate.

>>190096
Oh yeah im retarded. The voltage is the same across the parallel resistors, you're right

>>190096
Total voltage across the circuit is 46V. R1 is only one part of the circuit. So it makes sense that the voltage across R1 is less than the total voltage. The voltage across the parallel part would be the other component of the total voltage. Add these two up and you get the total

Let me start from the totals first, With Vs as 46 V, and Rt at 92 Ohms, We can calculate The Current by using V/R which is 46/92, resulting in .5A. I can then use this to find the voltage drop for R1 because series has the same current, so I can then do .5A*68 Ohms, correct? But I get 34 V as the drop, which looks incorrect to me.

>>190109
It's correct. Why does it look wrong to you?

File: ohgodwhy.png (342KB, 500x500px) Image search: [Google]

342KB, 500x500px

>>190113
Denial I guess, I just find it weird since it's only 1 resistor with that much drop. If it's correct, then the rest of the resistors are 12 V, and I can then find the current for each, and then the power as well. Gonna calculate and then post a picture.

I appreciate the help you anons gave me, it really helped clear some stuff up.

This is how I was feeling before lol. pic related

File: image.jpg (6MB, 4032x3024px) Image search: [Google]

6MB, 4032x3024px

So really the only values that should add up to the total are the current and power, the other two are weird.

>>190130
Thank you guys again, the second part should be a breeze now, really sorry for the picture again, I don't know how to resolve that.

>>190132
Move out of Australia.

>>190196
It's the Iphone, man. I can't figure it out.

>>190198
>I don't want to read manuals, I will just keep apologizing