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JEE Mains · Physics · STD 11 - Units,dimensions and measurement
  • JEE Mains 2021
  • Hard
  • MCQ

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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Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\frac{1}{{R}_{{eq}}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}\)

\(\frac{1}{{R}_{{eq}}}=\frac{1}{4}+\frac{1}{4} \Rightarrow {R}_{{eq}}=2\, \Omega\)

Also \(\frac{\Delta {R}_{{eq}}}{{R}_{{eq}}^{2}}=\frac{\Delta {R}_{1}}{{R}_{1}^{2}}+\frac{\Delta {R}_{2}}{{R}_{2}^{2}}\)

\(\frac{\Delta {R}_{{eq}}}{4}=\frac{.8}{16}+\frac{.4}{16}=\frac{1.2}{16}\)

\(\triangleq {R}_{{eq}}=0.3 \,\Omega\)

\({R}_{{eq}}=(2 \pm 0.3)\, \Omega\)

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