If 400; Omega of resistance is made by adding | vedclass

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(B) When resistors are connected in series,the total resistance is $R_{eq} = R_1 + R_2 + R_3 + R_4 = 100 + 100 + 100 + 100 = 400\; \Omega$.For resisto

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$5$

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(B) When resistors are connected in series,the total resistance is $R_{eq} = R_1 + R_2 + R_3 + R_4 = 100 + 100 + 100 + 100 = 400\; \Omega$.For resistors in series,the absolute error in the total resistance is the sum of the absolute errors of individual resistors: $\Delta R_{eq} = \Delta R_1 + \Delta R_2 + \Delta R_3 + \Delta R_4$.Given that the tolerance is $5\%$,the absolute error for each $100\; \Omega$ resistor is $\Delta R = 5\% 	ext{ of } 100\; \Omega = 5\; \Omega$.Thus,$\Delta R_{eq} = 5 + 5 + 5 + 5 = 20\; \Omega$.The percentage tolerance of the combination is $\frac{\Delta R_{eq}}{R_{eq}} 	imes 100\% = \frac{20}{400} 	imes 100\% = 5\%$.

If $400\; \Omega$ of resistance is made by adding four $100\; \Omega$ resistors of tolerance $5\%$ each,then the tolerance of the combination is .....$\%$

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