When two resistors of resistances R_{1} and R | vedclass

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Question

(A) When $R_{1}$ and $R_{2}$ are connected in parallel,the net resistance $(R_{p})$ is given by:$\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$T

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$4 \, \Omega$ and $12 \, \Omega$

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(A) When $R_{1}$ and $R_{2}$ are connected in parallel,the net resistance $(R_{p})$ is given by:$\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$Therefore,we have:$R_{p} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = 3 \, \Omega$ $...(1)$When $R_{1}$ and $R_{2}$ are connected in series,the net resistance $(R_{s})$ is given by:$R_{s} = R_{1} + R_{2} = 16 \, \Omega$ $...(2)$Substituting equation $(2)$ into equation $(1)$:$\frac{R_{1} R_{2}}{16} = 3 \implies R_{1} R_{2} = 48$ $...(3)$From equation $(2)$,$R_{2} = 16 - R_{1}$. Substituting this into equation $(3)$:$R_{1}(16 - R_{1}) = 48$$16R_{1} - R_{1}^{2} = 48$$R_{1}^{2} - 16R_{1} + 48 = 0$$(R_{1} - 12)(R_{1} - 4) = 0$Thus,$R_{1} = 12 \, \Omega$ or $R_{1} = 4 \, \Omega$.If $R_{1} = 12 \, \Omega$,then $R_{2} = 4 \, \Omega$. If $R_{1} = 4 \, \Omega$,then $R_{2} = 12 \, \Omega$.Therefore,the resistances are $4 \, \Omega$ and $12 \, \Omega$.

When two resistors of resistances $R_{1}$ and $R_{2}$ are connected in parallel,the net resistance is $3 \, \Omega$. When connected in series,its value is $16 \, \Omega$. Calculate the values of $R_{1}$ and $R_{2}$.

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