Example of a circuit partially in series and partially parallel
Voltage difference or electromotive force (emf) of the battery = 9.0 V
R3 = 3.0 Ω ; R1 = 4.0 Ω ; R2= 6.0 Ω
R1 en R2 are parallel and this is entirely in serie with R3

Calculate the current I1 in resistor R1
At first: find the equivalent resistance of R1 en R2
These two resistors are parallel.
1/R12 = 1/R1 + 1/R2
1/R12 = ¼ + 1/6 = 0.25 + 0.167 = 0.417
R12 = 1/0.417 = 2.40 Ω

The circuit is equivalent to :
RAC = 2.40 + 3.0 = 5.40 Ω
The main current :
V= IRAC
I = V/ RAC = 9.0/ 5.40 = 1.67 A
Calculate now VAB
VAB = IR12 = 1.67 x 2.40 = 4.0 V
Or :
VBC = I R3= 1.67 x 3.0 = 5.0 V
V = VAB + VBC
9.0 = VAB + 5.0
VAB = 4.0 V
VAB = I1 R1
4.0 = I1 x 40
I1 = 1.0 A
In the same way
VAB =I2R2
4.0 = I2 x 6.0
I2 = 0.67 A
Controle : I = I1 + I2