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