Answer Happy

You measure the voltage across a 5.3KΩ ± 1% resistor to be 2.397
V. Calculate the power dissipated by the resistor and
the absolute accuracy of you
calculation.
Accuracy x + Ax Ax is the absolute accuracy measurand accuracy of measurand Propagation of Accuracy for calculated values C is the calculated value and AC is the absolute accuracy of the calculated value Addition/Subtraction C = x₁ ± x₂ ± ... C = X₁ X₂ X3... Or Multiplication/Division C = x₁ + x₂ + x3 + ... Or mixed C = X₁ X₂ X3... Squares C = x₁(x₂)²(x₂)³.... 4x is the relative accuracy X (usually expressed as a percentage by multiplying by 100%) AC = √(Ax₂)² + (Ax₂)² + ... add absolute accuracy in quadrature 2 AC Ax₁ = + + + *** C add relative accuracy in quadrature 2 2 2 AC +4 +9 + www C X2 X3 add relative accuracy in quadrature weighted by exponent