Finding total resistance in a circuit with two parallel and one in series(due in a couple hours)

Ok, so am I doing this right?

Resistance of 220.0 and 100.0 are parallel, and resistance of 22.0 is in the series.

Is the total resistance 1/220 + 1/100 = 1/Rtotal, which then divides by one for 72 Ohms for the parallel, then add 20 Ohms so it equals 92 Ohms?

If so, then to find voltage and amps for each resistance, I should treat the system as a series now?

Please help. Thanks!

1/220 + 1/100 = 1/Rtotal

R total = 68.75Ω

Now 68.75 Ω and the 22 Ω are in series

Total = 90.75 Ω

Find the total current in the circuit I = (V / 90.75) ----------------1

P,D across the 22 Ω = current * resistance = I * 22
Uae equation 1 to find I



P.D across 220 Ω = P.D across 100 Ω = P,D across the parallel combination is current * resistance V ( parallel) = I * 68.75Ω

Current through 220 Ω = V ( parallel) / 220 Ω
Current through 100 Ω = V ( parallel)/ / 100 Ω

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yes, the resultant of parallel combination of 220 and 100 ohm is Rp = 275 / 4 = 68.75 ohm

and then the net resistance value is R = 68.75 + 22 = 90.75 ohm. so it is 91 ohm nearly not 92 ohm as you mentioned. then to find the values of voltage and current make use of Ohm's law V = IR