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(B) Let the three resistors be $R_1, R_2,$ and $R_3$. The equivalent resistance in parallel is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R

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$6$

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(B) Let the three resistors be $R_1, R_2,$ and $R_3$. The equivalent resistance in parallel is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = 1$.Given the ratio of two resistors is $R_1 : R_2 = 1 : 2$,we can write $R_2 = 2R_1$.Substituting this into the parallel resistance formula: $\frac{1}{R_1} + \frac{1}{2R_1} + \frac{1}{R_3} = 1$.This simplifies to $\frac{3}{2R_1} + \frac{1}{R_3} = 1$,or $\frac{1}{R_3} = 1 - \frac{3}{2R_1} = \frac{2R_1 - 3}{2R_1}$.Thus,$R_3 = \frac{2R_1}{2R_1 - 3}$.For $R_3$ to be a positive integer,$2R_1 - 3$ must be a divisor of $2R_1$. Since $2R_1 = (2R_1 - 3) + 3$,then $(2R_1 - 3)$ must be a divisor of $3$.The possible divisors of $3$ are $1$ and $3$.Case $1$: $2R_1 - 3 = 1 \Rightarrow 2R_1 = 4 \Rightarrow R_1 = 2 \ \Omega$. Then $R_2 = 2R_1 = 4 \ \Omega$ and $R_3 = \frac{4}{1} = 4 \ \Omega$. (Not unequal resistors).Case $2$: $2R_1 - 3 = 3 \Rightarrow 2R_1 = 6 \Rightarrow R_1 = 3 \ \Omega$. Then $R_2 = 2R_1 = 6 \ \Omega$ and $R_3 = \frac{6}{3} = 2 \ \Omega$.The resistors are $3 \ \Omega, 6 \ \Omega,$ and $2 \ \Omega$. The largest resistor is $6 \ \Omega$.

The equivalent resistance of three unequal resistors connected in parallel is $1 \ \Omega$. If the ratio of two of them is $1:2$ and all resistance values are integers,what is the value of the largest resistor in ohms?

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