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

circuit

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

Circuit

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              

  

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