Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32.
Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1 Ω resistance.
The resistance of each resistor connected in the given circuit, R = 1 Ω
Equivalent resistance of the given circuit = R’
The network is infinite. Hence, equivalent resistance is given by the relation,
$\therefore R^{\prime}=2+\frac{R^{\prime}}{\left(R^{\prime}+1\right)}$
$\left(R^{\prime}\right)^{2}-2 R^{\prime}-2=0$
$R^{\prime}=\frac{2 \pm \sqrt{4+8}}{2}$
$=\frac{2 \pm \sqrt{12}}{2}=1 \pm \sqrt{3}$
Negative value of R’ cannot be accepted. Hence, equivalent resistance,
$R^{\prime}=(1+\sqrt{3})=1+1.73=2.73 \Omega$
Internal resistance of the circuit, r = 0.5 Ω
Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23 Ω
Supply voltage, V = 12 V
According to Ohm's Law, current drawn from the source is given by the ratio, $\frac{12}{3.23}=3.72 \mathrm{~A}$
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