A 400 , Omega resistor is formed by connectin | vedclass

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(B) Given that, the resistance of each resistor is $R = 100 \, \Omega$ and the tolerance is $T = 5 \%$.When resistors are connected in series, the tot

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$5$

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(B) Given that, the resistance of each resistor is $R = 100 \, \Omega$ and the tolerance is $T = 5 \%$.When resistors are connected in series, the total resistance is $R_{eq} = R_1 + R_2 + R_3 + R_4 = 400 \, \Omega$.The absolute error in each resistor is $\Delta R = \frac{5}{100} 	imes 100 = 5 \, \Omega$.The total absolute error in the series combination is $\Delta R_{eq} = \Delta R_1 + \Delta R_2 + \Delta R_3 + \Delta R_4 = 5 + 5 + 5 + 5 = 20 \, \Omega$.The percentage tolerance of the combination is given by $T_{eq} = \left( \frac{\Delta R_{eq}}{R_{eq}} ight) 	imes 100 \%$.$T_{eq} = \left( \frac{20}{400} ight) 	imes 100 \% = 5 \%$.

$A$ $400 \, \Omega$ resistor is formed by connecting four $100 \, \Omega$ resistors, each having a tolerance of $5 \%$. What will be the tolerance of this combination in $\%$?

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