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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
$(4 \pm 0.3) \,\Omega$
(JEE MAIN-2021)
Solution
$\frac{1}{{R}_{{eq}}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}$
$\frac{1}{{R}_{{eq}}}=\frac{1}{4}+\frac{1}{4} \Rightarrow {R}_{{eq}}=2\, \Omega$
Also $\frac{\Delta {R}_{{eq}}}{{R}_{{eq}}^{2}}=\frac{\Delta {R}_{1}}{{R}_{1}^{2}}+\frac{\Delta {R}_{2}}{{R}_{2}^{2}}$
$\frac{\Delta {R}_{{eq}}}{4}=\frac{.8}{16}+\frac{.4}{16}=\frac{1.2}{16}$
$\triangleq {R}_{{eq}}=0.3 \,\Omega$
${R}_{{eq}}=(2 \pm 0.3)\, \Omega$
Standard 11
Physics
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