Two resistors of resistances R_1 = (100 pm 3) | vedclass

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(A) For resistors connected in series,the equivalent resistance is $R = R_1 + R_2$.$R = 100 \,\Omega + 200 \,\Omega = 300 \,\Omega$.The maximum absolu

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$7 \,\Omega, 2.3 \%$

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(A) For resistors connected in series,the equivalent resistance is $R = R_1 + R_2$.$R = 100 \,\Omega + 200 \,\Omega = 300 \,\Omega$.The maximum absolute error in the series combination is the sum of the individual absolute errors: $\Delta R = \Delta R_1 + \Delta R_2 = 3 \,\Omega + 4 \,\Omega = 7 \,\Omega$.Thus,the equivalent resistance is $(300 \pm 7) \,\Omega$.The percentage error is given by $\frac{\Delta R}{R} 	imes 100 \%$.Percentage error $= \frac{7}{300} 	imes 100 \% = \frac{7}{3} \% \approx 2.33 \%$.Rounding to one decimal place,we get $2.3 \%$.

Two resistors of resistances $R_1 = (100 \pm 3) \,\Omega$ and $R_2 = (200 \pm 4) \,\Omega$ are connected in series. The maximum absolute error and percentage error in the equivalent resistance of the series combination are:

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