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

Q: 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

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

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

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, Diffcult

STD 11 Science
JEE
STD 11 - 1. units,dimensions and measurement
Physics

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