The values of two resistors are $R_{1} = (6 \pm 0.3)k\Omega$ and $R_{2} = (10 \pm 0.2)k\Omega$. The percentage error in the equivalent resistance when they are connected in parallel is

Question

Dr. Abubacker · Accepted Answer

$R_{parallel} = \frac{R_{1}R_{2}}{(R_{1} + R_{2})}$ $\Rightarrow \frac{\mathrm{\Delta}R_{p}}{R_{p}} = \frac{\mathrm{\Delta}R_{1}}{R_{1}} + \frac{\mathrm{\Delta}R_{2}}{R_{2}} + \frac{\mathrm{\Delta}(R_{1} + R_{2})}{R_{1} + R_{2}}$ $\Rightarrow \frac{\mathrm{\Delta}\ R_{p}}{R_{p}} = \frac{0.3}{6} + \frac{0.2}{10} + \frac{(0.3 + 0.2)}{10 + 6}$ $= 0.05 + 0.02 + 0.03125 = 0.10125$ $	herefore\frac{\mathrm{\Delta}R_{p}}{R_{p}} 	imes 100 = 10.125$ or $10.125\%$

Q. The values of two resistors are $R_{1} = (6 \pm 0.3)k\Omega$ and $R_{2} = (10 \pm 0.2)k\Omega$. The percentage error in the equivalent resistance when they are connected in parallel is

Explanation

Reviewed by Dr. Abubacker

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