# 2 x 10 3 a 95 then since voltage e fir 1 2 2 2 e

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Unformatted text preview: mA Accuracy: within 0.100% of reading FIND: (ud)E SOLUTION From Ohm's Law: E = IR or in terms of power, P =I2R. For the nominal values of power and resistance given, expect a current, I = (P/R)1/2 = 4.08 A. Hence, Ohmeter (ud)R= (uo2 + uc2)1/2 = ((0.00500 x 30.0 Ω )2+ (0.500 Ω )2)1/2 = 0.522 Ω (95%) Ammeter (ud)I = (uo2 + uc2)1/2 = ((50.0 x 10-3 A)2+ (0.00100 x 4.08 A)2)1/2 = 50.2 x 10-3 A (95%) Then, since voltage E = f(I,R): 1/ 2 2 2 ⎡ྎ⎛ྎ ∂E ⎞ྏ ⎛ྎ ∂E ⎞ྏ ⎤ྏ = ± ⎡ྎ R(u ) (u d )E = ± ⎢ྎ⎜ྎ (u d )I ⎟ྏ + ⎜ྎ (u ) ⎟ྏ ⎥ྏ dI ⎢ྎ ⎣ྏ ⎠ྏ ⎝ྎ ∂R d R ⎠ྏ ⎥ྏ ⎢ྎ⎝ྎ ∂I ⎣ྏ ⎦ྏ ( [ 2 = ± ( 30Ω × 0.0502 A) + ( 4.08...
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## This note was uploaded on 01/07/2013 for the course ENGR 201 taught by Professor Miller during the Spring '08 term at Drexel.

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