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(C) Let the resultant resistance between $A$ and $B$ be $R$. Since the ladder is infinite,adding one more section of $R_1$ and $R_2$ to the existing i

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$2$

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(C) Let the resultant resistance between $A$ and $B$ be $R$. Since the ladder is infinite,adding one more section of $R_1$ and $R_2$ to the existing infinite network will not change the total resistance $R$.The circuit can be simplified as a resistor $R_1$ in series with the parallel combination of $R_2$ and $R$.Thus,the equivalent resistance is given by:$R = R_1 + \frac{R \cdot R_2}{R + R_2}$Substituting the given values $R_1 = 1\,\Omega$ and $R_2 = 2\,\Omega$:$R = 1 + \frac{2R}{R + 2}$Multiplying both sides by $(R + 2)$:$R(R + 2) = 1(R + 2) + 2R$$R^2 + 2R = R + 2 + 2R$$R^2 - R - 2 = 0$Solving the quadratic equation:$(R - 2)(R + 1) = 0$Since resistance cannot be negative,we have $R = 2\,\Omega$.

An infinite sequence of resistors is shown in the figure. The resultant resistance between $A$ and $B$ will be,when $R_1 = 1\,\Omega$ and $R_2 = 2\,\Omega$ ............. $\Omega$

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