Two resistors R₁=(4±0.8)Ω and R₂... | Errors

JEE MAIN 2021

01-09-21 S2

Question

Two resistors $\mathrm{R}_1=(4 \pm 0.8) \Omega$ and $\mathrm{R}_2=(4 \pm 0.4) \Omega$ are connected in parallel. The equivalent resistance of their parallel combination will be :

Select the correct option:

$(4 \pm 0.4) \Omega$

$(2 \pm 0.4) \Omega$

$(2 \pm 0.3) \Omega$

$(4 \pm 0.3) \Omega$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$$ \begin{aligned} & \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2} \\ & \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{4}+\frac{1}{4} \Rightarrow \mathrm{R}_{\mathrm{eq}}=2 \Omega \end{aligned} $$ Also $$ \frac{\Delta \mathrm{R}_{\text {eq }}}{\mathrm{R}_{\text {eq }}}=\frac{\Delta \mathrm{R}_1}{\mathrm{R}_1^2}+\frac{\Delta \mathrm{R}_2}{\mathrm{R}_2^2} $$ $$ \frac{\Delta \mathrm{R}_{\mathrm{eq}}}{4}=\frac{.8}{16}+\frac{.4}{16}=\frac{1.2}{16} $$ $$ \begin{aligned} & \Delta \mathrm{R}_{\mathrm{eq}}=0.3 \Omega \\ & \mathrm{R}_{\mathrm{eq}}=(2 \pm 0.3) \Omega \end{aligned} $$ Option (3)

Question Tags

JEE Main

Physics

Easy

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