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(B) Let the three resistors be $x$,$2x$,and $R$. Since they are connected in parallel,their equivalent resistance is given by $\frac{1}{R_{eq}} = \fra

Vedclass · Accepted Answer

$6$

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(B) Let the three resistors be $x$,$2x$,and $R$. Since they are connected in parallel,their equivalent resistance is given by $\frac{1}{R_{eq}} = \frac{1}{x} + \frac{1}{2x} + \frac{1}{R}$.Given $R_{eq} = 1 \, \Omega$,we have $\frac{1}{x} + \frac{1}{2x} + \frac{1}{R} = 1$.Simplifying the expression: $\frac{3}{2x} + \frac{1}{R} = 1 \Rightarrow \frac{1}{R} = 1 - \frac{3}{2x} = \frac{2x - 3}{2x}$.Thus,$R = \frac{2x}{2x - 3}$.Since $R$ must be an integer (non-fractional),$2x - 3$ must be a divisor of $2x$. We can rewrite $R = \frac{(2x - 3) + 3}{2x - 3} = 1 + \frac{3}{2x - 3}$.For $R$ to be an integer,$2x - 3$ must be a factor of $3$. The factors of $3$ are $1$ and $3$.Case $1$: $2x - 3 = 1 \Rightarrow 2x = 4 \Rightarrow x = 2$. Then $R = 1 + \frac{3}{1} = 4$. The resistors are $2, 4, 4$. (Rejected,as resistors must be unequal).Case $2$: $2x - 3 = 3 \Rightarrow 2x = 6 \Rightarrow x = 3$. Then $R = 1 + \frac{3}{3} = 2$. The resistors are $3, 6, 2$. Checking the values: $\frac{1}{3} + \frac{1}{6} + \frac{1}{2} = \frac{2+1+3}{6} = 1$. This satisfies the condition.The resistances are $2 \, \Omega, 3 \, \Omega, 6 \, \Omega$. The largest resistance is $6 \, \Omega$.

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

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