solve this


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:

  1. $(4 \pm 0.4) \Omega$

  2. $(2 \pm 0.4) \Omega$

  3. $(2 \pm 0.3) \Omega$

  4. $(4 \pm 0.3) \Omega$

Correct Option: , 3


$\frac{1}{R_{\text {cq }}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$

$\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{4}+\frac{1}{4} \Rightarrow \mathrm{R}_{\mathrm{eq}}=2 \Omega$

Also $\frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}^{2}}=\frac{\Delta \mathrm{R}_{1}}{\mathrm{R}_{1}^{2}}+\frac{\Delta \mathrm{R}_{2}}{\mathrm{R}_{2}^{2}}$

$\frac{\Delta \mathrm{R}_{\mathrm{eq}}}{4}=\frac{.8}{16}+\frac{.4}{16}=\frac{1.2}{16}$

$\underline{\underline{\Delta}} \mathrm{R}_{\mathrm{eq}}=0.3 \Omega$

$\mathrm{R}_{\mathrm{eq}}=(2 \pm 0.3) \Omega$

Option (3)

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