The load across a 120-V power supply consists of three resistances are R1 = 10 Ω, R2 = 30 Ω, and R3 is unknown. If R2 has a difference of potential of 45 V across it, find the other voltage drops across the other resistors, the current through each resistor, and R3.

Given R2’s resistance and voltage we can find the current through it of 1.5 amps. Due to Kirchhoff’s junction rule the current going in must match the current going out so the current through them all is 1.5 amps. Using this we can find the voltage through R1 of 15v. Then we subtract V1+V2 from 120 to find that R3 has a voltage of 60v. Next we find that R3 has a resistance of 40 ohms