I\'m stuck on a problem dealing with relative uncertainty, the problem says: An

Question

I'm stuck on a problem dealing with relative uncertainty, the problem says:
An ammeter and an ohmmeter are used to indirectly measure the electrical power
dissipated in a resistor.
a. Show by analysis that if the relative uncertainty in the direct current
measurement, (?I/I), is +1%, and the relative uncertainty in the direct resistance
measurement, (?R/R), is +2%, that the maximum relative uncertainty in the
indirectly measured power, (?P/P), will be +4%.
b. If the nominal power, P, is 100W, what is the power uncertainty, ?P, in watts?
Note: P = RI2. Textbook: Introduction to Thermal Dynamics and Heat Transfer 2nd edition, Yunus Cengel
Note: P = RI2. Textbook: Introduction to Thermal Dynamics and Heat Transfer 2nd edition, Yunus Cengel

Explanation / Answer

(a) uncertainity in the current measurement = +1%
uncertainity in the resistance measurement = +2%
we know power, P = I2R ?uncertainity in the power measurement is P = (+2)2 x (+1)    = +4% (b)Given, P = 100W uncertainity in the power = +4% ?power uncertainity in Watts = 100 + 100 x (4/100)                                           = 104 Watts

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---