Series/Parallel circuit

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 :

R_AC  = 2.40 + 3.0 = 5.40 Ω     

The main current :

V= IR_AC

I = V/ R_AC  = 9.0/ 5.40 = 1.67 A

Calculate now  V_AB

V_AB = IR₁₂ = 1.67 x 2.40 = 4.0 V          

Or :

V_BC = I R₃= 1.67 x 3.0 = 5.0 V 

V ₌ V_AB + V_BC  

9.0 = V_AB + 5.0

V_AB = 4.0 V 

V_AB = I₁ R₁

 4.0 = I₁  x 40

 I₁ = 1.0 A

In the same way

V_AB =I₂R₂

4.0 = I₂ x 6.0

I₂ = 0.67 A

Controle : I = I₁ + I₂