The equivalent resistance of two resistors co | vedclass

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Question

(C) Let the two resistances be $R_{1}$ and $R_{2}$.When they are connected in series,the equivalent resistance is $R_{s} = R_{1} + R_{2} = 6 \ \Omega$

Vedclass · Accepted Answer

$4 \ \Omega, 2 \ \Omega$

Vedclass · Answer

(C) Let the two resistances be $R_{1}$ and $R_{2}$.When they are connected in series,the equivalent resistance is $R_{s} = R_{1} + R_{2} = 6 \ \Omega$ $(1)$.When they are connected in parallel,the equivalent resistance is $R_{p} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{4}{3} \ \Omega$ $(2)$.Substituting $R_{1} + R_{2} = 6$ into equation $(2)$:$\frac{R_{1} R_{2}}{6} = \frac{4}{3} \Rightarrow R_{1} R_{2} = 6 	imes \frac{4}{3} = 8 \ \Omega^{2}$ $(3)$.From $(1)$,$R_{2} = 6 - R_{1}$. Substituting this into $(3)$:$R_{1}(6 - R_{1}) = 8 \Rightarrow 6R_{1} - R_{1}^{2} = 8 \Rightarrow R_{1}^{2} - 6R_{1} + 8 = 0$.Solving the quadratic equation:$(R_{1} - 4)(R_{1} - 2) = 0$.Thus,$R_{1} = 4 \ \Omega$ or $R_{1} = 2 \ \Omega$.If $R_{1} = 4 \ \Omega$,then $R_{2} = 2 \ \Omega$. If $R_{1} = 2 \ \Omega$,then $R_{2} = 4 \ \Omega$.Therefore,the two resistances are $4 \ \Omega$ and $2 \ \Omega$.

The equivalent resistance of two resistors connected in series is $6 \ \Omega$ and their parallel equivalent resistance is $\frac{4}{3} \ \Omega$. What are the values of the resistances?

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