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Numerical · Q23

Q.Two resistors, R1=(4.0±0.2) ΩR_1 = (4.0 \pm 0.2)\ \Omega and R2=(6.0±0.3) ΩR_2 = (6.0 \pm 0.3)\ \Omega, are connected in series. Find the equivalent resistance together with its absolute error.

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For resistors in series, the equivalent resistance is the simple sum R=R1+R2R = R_1 + R_2:

R=4.0 Ω+6.0 Ω=10.0 ΩR = 4.0\ \Omega + 6.0\ \Omega = 10.0\ \Omega

For a sum (or difference) of two measured quantities, the combined-error rule states that the absolute errors add, regardless of whether the quantities themselves are being added or subtracted: …

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