Determine the current drawn from a 12V supply with internalresistance 0.5W by the infinite network shown in Fig. 3.32. Each resistor has 1W resistance.
Consider the given network shown in the figure below.
The equivalent resistance for the infinite network is given as,
R ′ = 2 + R ′ R ′ + 1
Where, the equivalent resistance for the infinite network is R ′ .
By simplifying the above expression, we get
( R ′ ) 2 − 2 R ′ − 2 = 0 R ′ = 2 ± 4 + 8 2 R ′ = 1 ± 3
Since, the resistance cannot have negative value therefore,
R ′ = 1 + 3 = 2.73 Ω
The current drawn from the supply is given as,
I = V r + R ′
Where, the supply voltage is V and the internal resistance is r .
By substituting the given values in the above expression, we get
I = 12 0.5 + 2.73 = 12 3.23 = 3.72 A
Thus, the current drawn from the supply is 3.72 A .
Consider the given network shown in the figure below.
The equivalent resistance for the infinite network is given as,
Where, the equivalent resistance for the infinite network is
By simplifying the above expression, we get
Since, the resistance cannot have negative value therefore,
The current drawn from the supply is given as,
Where, the supply voltage is
By substituting the given values in the above expression, we get
Thus, the current drawn from the supply is
