Question

Error Calculation Let R, R2, R3 denote the resistances of three resistors in parallel. It follows that the total resistance, R, satisfies: have listed values of 202, respectively all with an error of at Suppose RiR2, R most 5%. 302, and 502 (a) Find an approximate bound for the error of the total resistance R. (b) Suppose you were considering upgrading one of the resistors to have an error of at most 1%. Which resistor should be upgraded in order to reduce the error in R the most? Show work in support of your answer

Answer 1

1/R = 1/R_1 + 1/R_2 + 1/R_3 R_1 = 20 ohm plusminus 5% = 20 ohm plusminus 1 ohm R_2 = 30 ohm plusminus 5% = 30 ohm plusminus 1.5 ohm R_3 = 50 ohm plusminus 5% = 50 ohm plusminus 2.5 ohm delta R/R^2 = delta R_1 /R^2_1 + delta R_2 /R^2_2 + delta R_3 /R^2_3 here R = 1/1/R_1 + 1/R_2 + 1/R_3 = 1/1/20 + 1/30 + 1/50 R = 9.6774 ohm so delta R/(9.6774)^2 = 1/20^2 + 1.5/30^2 + 2.5/50^2 delta R = 0.4838 ohm so R = 9.6774 ohm plusminus 0.4838 R = 9.677a plusminus 5% from expression (1) , If we take R_1 R_1 = 20 ohm plusminus 1% Become delta R/(9.6771)^2 = 1/100 times 20/(20)^2 + 1.5/(30)^2 + 2.5/(50)^2 \r\n delta R = 0.2965 ohm So R = 9.6774 plusminus 0.2965 ohm R = 9.6774 plusminus 3% so R_1 Impact is high compared to others due to R_1 is less value compared to