In a simple electrical circuit,the current in a resistor is measured as (2.50 0.05) mA.T
问题描述：
In a simple electrical circuit,the current in a resistor is measured as (2.50 0.05) mA.T
he resistor is marked as having a value of 4.7欧姆 正负 2%.
If these values were used to calculate the power dissipated in the resistor using the equation
P = I2R,what would be the percentage uncertainty in the value obtained?
答
the percentage uncertainty up/p=P'*[(2ΔI/I)^2+(ΔR/R)^2]^(1/2)
=2.50^2*4.7*[(2*0.05/2.50)^2+0.02^2]^(1/2)
=29.375*20^(1/2)/100
=1.3%