Two resistances r_1 and r_2 (r_1

Question

(D) When two resistors $r_1$ and $r_2$ are connected in parallel, the equivalent resistance $R$ is given by the formula: $\frac{1}{R} = \frac{1}{r_1}

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$R < r_1$

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(D) When two resistors $r_1$ and $r_2$ are connected in parallel, the equivalent resistance $R$ is given by the formula: $\frac{1}{R} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2}$.Therefore, $R = \frac{r_1 r_2}{r_1 + r_2}$.Given that $r_1 Since $R = \frac{r_1 r_2}{r_1 + r_2}$, we can write $R = r_1 \left( \frac{r_2}{r_1 + r_2} \right)$.Since $r_2 Thus, $R Similarly, it can be shown that $R < r_2$. Therefore, the equivalent resistance in a parallel combination is always less than the smallest individual resistance in the circuit.

Two resistances $r_1$ and $r_2$ $(r_1 < r_2)$ are joined in parallel. The equivalent resistance $R$ is such that

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