Two resistors R_{1} = (4 pm 0.8) , Omega and | vedclass

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(C) For two resistors connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$.Subs

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$(2 \pm 0.3) \, \Omega$

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(C) For two resistors connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$.Substituting the given values,$\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$,which gives $R_{eq} = 2 \, \Omega$.To find the error $\Delta R_{eq}$,we differentiate the expression $\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$:$-\frac{\Delta R_{eq}}{R_{eq}^{2}} = -\frac{\Delta R_{1}}{R_{1}^{2}} - \frac{\Delta R_{2}}{R_{2}^{2}}$.Taking the magnitude of errors,$\frac{\Delta R_{eq}}{R_{eq}^{2}} = \frac{\Delta R_{1}}{R_{1}^{2}} + \frac{\Delta R_{2}}{R_{2}^{2}}$.Substituting the values: $\frac{\Delta R_{eq}}{2^{2}} = \frac{0.8}{4^{2}} + \frac{0.4}{4^{2}}$.$\frac{\Delta R_{eq}}{4} = \frac{0.8 + 0.4}{16} = \frac{1.2}{16}$.$\Delta R_{eq} = 4 	imes \frac{1.2}{16} = \frac{1.2}{4} = 0.3 \, \Omega$.Therefore,the equivalent resistance is $R_{eq} = (2 \pm 0.3) \, \Omega$.

Two resistors $R_{1} = (4 \pm 0.8) \, \Omega$ and $R_{2} = (4 \pm 0.4) \, \Omega$ are connected in parallel. The equivalent resistance of their parallel combination will be

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