Current Divider Circuit Numericals

For the given circuit calculate the value of the current in either branch and the value of the unknown resistance R when the total current taken by the network is 2.25 A.

Solution
The circuit diagram for above problem is drawn below :-
Applied voltage, V = 10 V
Circuit current, I = 2.25 A
Circuit resistance, R_eq = V/I = 10/2.25 = 4.44Ω
From the figure,

4.44Ω = (2Ω + 8Ω) || (5 + R)


4.44Ω = 10Ω || (5 + R)


4.44 =          1         
4.44 =                    
4.44 =   1/10 + 1/(5 + R) 


Or, 4.44 =              1              
Or, 4.44 =                             
Or, 4.44 =   (5 + R + 10) / 10(R + 5)  


Or, 4.44   10(R + 5)   
Or, 4.44               
Or, 4.44   5 + R + 10  


Or, 4.44(15 + R) = 10R + 50


Or, 66.67 + 4.44R = 10R + 50


Or, 66.67 - 50 = 10R - 4.44R


Or, 16.67 = 5.56R


Or, R = 16.67/5.56 = 2.997Ω  ≃ 3Ω

The value of unknown resistance is found to be 2.997Ω or approximately 3Ω.

Let current "I₁" flows through the branch of 2Ω and 8Ω resistor and "2.25 - I₁" flows through the branch of 5Ω and 3Ω resistors.

Applying current divider rule :-
Current flowing through branch is inversely proportional to the resistance path.

     I₁       =      8  
              =         
  2.25 - I₁   =      10 

Or, 10I₁ = 18 - 8I₁
Or, 10I₁ + 8I₁ = 18
Or, 18I₁ = 18
∴ I₁ = 1 A.
This I₁ = 1A flows through 2Ω and 8Ω resistor.
Current flowing through 5Ω and 3Ω resistors = 2.25 - I₁ = 2.25 - 1 = 1.25A Redrawing the circuit with value of R and branch current :-