Error Analysis. For a given coil, R_L ~ 0, so that, from Eq. (5.70) we have (V_L

Question

Error Analysis. For a given coil, R_L ~ 0, so that, from Eq. (5.70) we have (V_L/V_R) = (omega L/R) for this coil. This expression can be rearranged follows: L = V_L R/V_R omega. You can determine L for the coil described in this problem by measuring V_L, V_R, omega, and R, and by using Eq. (5.10). Assume that Delta omega = 0, and using the formula for error propagation, express the fractional uncertainty, (Delta L/L), in terms of the fractional uncertainties, (Delta V_L/V_L), (Delta V_R/V_R) and (Delta R/R).

Explanation / Answer

Here it is given that L= vL*R/[vR*w]

                 taking log of either side,

   ln L= ln vL*R/[vR*w]

differentiating each side,

deltaL/L = delta vL/vL + deltaR /R - deltavR/vR - deltaW/W

Since error always add up, and deltaW= 0;

deltaL/L = delta vL/vL + deltaR /R  + deltavR/vR    Answer

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