Three unequal resistances are connected in pa | vedclass

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(D) Let the three resistances be $R_1, R_2$,and $R_3$. The equivalent resistance $R_{	ext{eq}}$ for parallel connection is given by $\frac{1}{R_{	ex

Vedclass · Accepted Answer

$6$

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(D) Let the three resistances be $R_1, R_2$,and $R_3$. The equivalent resistance $R_{	ext{eq}}$ for parallel connection is given by $\frac{1}{R_{	ext{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.Given $R_{	ext{eq}} = 1 \Omega$,so $\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = 1$.Let the ratio of two resistances be $R_1 : R_2 = 1 : 2$,which implies $R_2 = 2R_1$.Substituting this into the equation: $\frac{1}{R_1} + \frac{1}{2R_1} + \frac{1}{R_3} = 1 \Rightarrow \frac{3}{2R_1} + \frac{1}{R_3} = 1$.We need to find integer values for $R_1, R_2, R_3$ such that they are unequal and satisfy the equation.If $R_1 = 2$,then $R_2 = 4$. Substituting: $\frac{3}{4} + \frac{1}{R_3} = 1 \Rightarrow \frac{1}{R_3} = \frac{1}{4} \Rightarrow R_3 = 4$. This makes $R_2 = R_3$,which contradicts the condition that resistances are unequal.If $R_1 = 3$,then $R_2 = 6$. Substituting: $\frac{3}{6} + \frac{1}{R_3} = 1 \Rightarrow \frac{1}{2} + \frac{1}{R_3} = 1 \Rightarrow \frac{1}{R_3} = \frac{1}{2} \Rightarrow R_3 = 2$. The resistances are $3 \Omega, 6 \Omega, 2 \Omega$. These are unequal and satisfy the condition.The values are $2 \Omega, 3 \Omega, 6 \Omega$. The highest resistance is $6 \Omega$.

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