Finding values in a series and parallel

You have 2 resistors (R1=220., R2=100.) that are in parallel with each other. These 2 are in series with a third resistor (R3=22.0). These resistors are in a circuit with a 5.00 V DC power supply . From this information, find Vtot, Req, Itot, I1, I2, I3, V1, V2 and V3. Show all of your work including units.

10 points for answer for sure. Thanks!

V total = source = 5 Volt dc
i total = 5V / [(220 ll100)Ohm + 22 Ohm = 55.1mA
i of R1 = (100/320)*(55.1mA) = 17.22mA
i of R2 = (220/320)*(55.1mA) = 37.88mA
i of R3 = i total = 55.1mA
V1 = (.01722A)*(220 Ohm) = 3.788 Volts
V2 = (.03788A)*(100 Ohm) = 3.788 Volts
V3 = (.0551A)*(22 Ohm) = 1.212 Volts

The symbol didn't show. I'll assume it was the ohm symbol (omega).

Use the power supply negative terminal as the reference point. Then:
Vtot = 5.00VDC <==ANSWER1

R1||R2 = 1/(1/(220ohms)+1/(100ohms))) = 68.75 ohms
Req = R1||R2+R3 = 68.75ohms + 22ohms = 90.75ohms
= 90.8 ohms (rounded) <==ANSWER2

Itot = Vtot/Req = 5.00VDC / 90.75ohms = 0.05510A
= 55.1mA (rounded) <==ANSWER3

I1 = V12/R1 = (ItotxR1||R2)/R1 = Itot x (R1||R2)/R1
= Itot x 1/(1/R1+1/R2) / R1 = Itot x 1/(1+R1/R2)
= Itot x R2/(R1+R2)
= 55.10mA x 100ohms/(220ohms+100ohms) = 17.22mA
= 17.2mA (rounded) <==ANSWER4

I2 = Itot x R1/(R1+R2) = 55.10mA X 220ohms/(220ohms+100ohms)
= 37.88mA = 37.9mA (rounded) <==ANSWER5

V1 = I1R1 = 0.01722A x 220ohm = 3.788V
= 3.79V (rounded) <==ANSWER6

V2 = I2R2 = 0.03788mA x 100ohm = 3.788V
= 3.79V (rounded) <==ANSWER7

V3 = I3R3 = Itot x R3 = 0.05510A x 22ohm = 1.212V
= 1.21V (rounded) <==ANSWER8