JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
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
- A \((4 \pm 0.4)\, \Omega\)
- B \((2 \pm 0.4)\, \Omega\)
- C \((2 \pm 0.3) \,\Omega\)
- D \((4 \pm 0.3) \,\Omega\)
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 \(\frac{\Delta {R}_{{eq}}}{{R}_{{eq}}^{2}}=\frac{\Delta {R}_{1}}{{R}_{1}^{2}}+\frac{\Delta {R}_{2}}{{R}_{2}^{2}}\)…
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