Two resistors of resistance R_1 and R_2 havin | vedclass

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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} +

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$R < R_2$

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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: $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$.This simplifies to $R = \frac{R_1 R_2}{R_1 + R_2}$.Given that $R_1 > R_2$,we can analyze the relationship between $R$ and the individual resistances.Since $R = \frac{R_1 R_2}{R_1 + R_2}$,we can write $R = R_2 \left( \frac{R_1}{R_1 + R_2} \right)$.Since $\frac{R_1}{R_1 + R_2} Similarly,since $R_2 Therefore,the equivalent resistance of a parallel combination is always less than the smallest individual resistance in the circuit. Thus,$R < R_2$ is the correct statement.

Two resistors of resistance $R_1$ and $R_2$ having $R_1 > R_2$ are connected in parallel. For equivalent resistance $R$,the correct statement is

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