Two resistors {R}_{1}=(4 pm 0.8) Omega and {R | vedclass

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Question

$\frac{1}{{R}_{{eq}}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}$

$\frac{1}{{R}_{{eq}}}=\frac{1}{4}+\frac{1}{4} \Rightarrow {R}_{{eq}}=2\, \Omega$

Also

Vedclass · Accepted Answer

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

Vedclass · Answer

$\frac{1}{{R}_{{eq}}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}$

$\frac{1}{{R}_{{eq}}}=\frac{1}{4}+\frac{1}{4} \Rightarrow {R}_{{eq}}=2\, \Omega$

Also $\frac{\Delta {R}_{{eq}}}{{R}_{{eq}}^{2}}=\frac{\Delta {R}_{1}}{{R}_{1}^{2}}+\frac{\Delta {R}_{2}}{{R}_{2}^{2}}$

$\frac{\Delta {R}_{{eq}}}{4}=\frac{.8}{16}+\frac{.4}{16}=\frac{1.2}{16}$

$	riangleq {R}_{{eq}}=0.3 \,\Omega$

${R}_{{eq}}=(2 \pm 0.3)\, \Omega$

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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