Three unequal resistors in parallel are equiv | vedclass

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(B) Let the three resistances be $R_1, R_2,$ and $R_3$.Given that $R_1:R_2 = 1:2$,we can write $R_1 = k$ and $R_2 = 2k$,where $k$ is a positive intege

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$6$

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(B) Let the three resistances be $R_1, R_2,$ and $R_3$.Given that $R_1:R_2 = 1:2$,we can write $R_1 = k$ and $R_2 = 2k$,where $k$ is a positive integer.For resistors in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.Given $R_{eq} = 1 \ \Omega$,we have $1 = \frac{1}{k} + \frac{1}{2k} + \frac{1}{R_3}$.Rearranging for $R_3$: $\frac{1}{R_3} = 1 - (\frac{1}{k} + \frac{1}{2k}) = 1 - \frac{3}{2k} = \frac{2k-3}{2k}$.Thus,$R_3 = \frac{2k}{2k-3}$.For $R_3$ to be a positive integer,$2k-3$ must be a divisor of $2k$. Since $\frac{2k}{2k-3} = \frac{2k-3+3}{2k-3} = 1 + \frac{3}{2k-3}$,$2k-3$ must be a divisor of $3$.The divisors of $3$ are $1$ and $3$.Case $1$: $2k-3 = 1 \Rightarrow 2k = 4 \Rightarrow k = 2$. Then $R_1 = 2, R_2 = 4, R_3 = 1 + \frac{3}{1} = 4$. Here $R_2 = R_3$,which violates the condition that resistors are unequal.Case $2$: $2k-3 = 3 \Rightarrow 2k = 6 \Rightarrow k = 3$. Then $R_1 = 3, R_2 = 6, R_3 = 1 + \frac{3}{3} = 2$. All resistances are unequal $(3, 6, 2)$.The largest resistance is $6 \ \Omega$.

Three unequal resistors in parallel are equivalent to a resistance of $1 \ \Omega$. If two of them are in the ratio $1:2$ and if no resistance value is fractional,then the largest of the three resistances in ohms is

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