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
Pick an option to check your answer.
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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