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Three unequal resistors in parallel are equivalent to a resistance $1 \mathrm{ohm}$. 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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6
Let the three resistances are $R_1, R_2$ and $R_3$.
$\therefore \quad \frac{R_1}{R_2}=\frac{1}{2} \Rightarrow R_1=k, R_2=2 k$
In parallel, $\quad \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$
$\frac{1}{1}=\frac{1}{k}+\frac{1}{2 k}+\frac{1}{R_3}$
$\frac{1}{R_3}=1-\frac{1}{k}-\frac{1}{2 k}$
$=\frac{2 k-2-1}{2 k}=\frac{2 k-3}{2 k}$
$R_3=\frac{2 k}{2 k-3}$
If $k=1$, then $R_3$ is found to be negative, which is impossible.
If $k=2$, then $R_1=2, R_2=4, R_3=4$
$R_2=R_3$, not satisfying the condition of the question that all resistance are unequal.
If $k=3$, then $R_1=3, R_2=6$
$R_3=2 \Omega$
$\therefore \quad$ Largest resistance $=6 \Omega$
$\therefore \quad \frac{R_1}{R_2}=\frac{1}{2} \Rightarrow R_1=k, R_2=2 k$
In parallel, $\quad \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$
$\frac{1}{1}=\frac{1}{k}+\frac{1}{2 k}+\frac{1}{R_3}$
$\frac{1}{R_3}=1-\frac{1}{k}-\frac{1}{2 k}$
$=\frac{2 k-2-1}{2 k}=\frac{2 k-3}{2 k}$
$R_3=\frac{2 k}{2 k-3}$
If $k=1$, then $R_3$ is found to be negative, which is impossible.
If $k=2$, then $R_1=2, R_2=4, R_3=4$
$R_2=R_3$, not satisfying the condition of the question that all resistance are unequal.
If $k=3$, then $R_1=3, R_2=6$
$R_3=2 \Omega$
$\therefore \quad$ Largest resistance $=6 \Omega$
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