1. The problem statement, all variables and given/known data In the circuit the emf is 150V connected across a resistance network. Each resistor has a value of 10 ohms. What is the current through R2? 2. Relevant equations R=V/I 3. The attempt at a solution I found that the resistance of R2 and R3 because they're in a series is 20 ohms. And because R1 is parallel to R2 and R3 i found the resistance to be 6.67 ohms. So the total resistance of the circuit would be 16.67 ohms including R4. And then I found that the total current drawn from the emf would be 8.998A. Now I'm just not sure how to isolate R2's current.
using the resistivity of R2 and R3 i multiplied it by the total current to find the potential drop and subtracted it from the emf voltage
But the total current doesn't go through R2 and R3. (If it did, then you'd already have your answer!) But it does go through R4.
So the current through R4 would be the sum of the current through R1, R2 and R3? So could i just divide the current of 8.998 into 2? So that R1 will have a current of 4.499 and then divide 4.499 into 2 again so that R2 and R3 will have a current of 2.2495 each?
The total current flows through R4, which is the sum of the currents through the two parallel branches (R1 and R2+R3). Note that R2 and R3 are in series, thus the same current passes through each. Find the voltage drop across R4, then use that to find the voltage across the R2+R3 branch.
Okay, so for R4 I get a voltage drop of 89.98 so voltage across R2+R3 would be 60.02? And then if i divide that by the total resistivity of R2+R3 i get 3.001, and since there's two i find that each resister will have 1.5 resistance?
Right. (Actually, if you used more digits in your calculation you'd get 60 V.) Right. 3 A passes through the R2+R3 branch. No. Those two resistors are in series, so they each get the full current through the branch.
Oh i see so because they're in a series they each have 3 going through them, but the current coming out at the end is still 3?