In a simple electric circuit, the current in resistor is measured as (2.50+_0.05)ma. the resistor is marked as having a value of 4.7ohm+_2 %
if these values were used to calculate the power dissipated in the in the resistor . what would be the percentage uncertainty in the value.

6%

Explanation:

current I = 2.5±.05 mA =2.5± .05x 100/2.5 = 2.5 ± 2%

resistance R =4.7 ± 2 %

power P = I² R

ΔP/Px 100 = 2ΔI/I x 100 + ΔR/R x 100

% uncertainty in P = 2x % uncertainty in I  + % uncertainty in R

% uncertainty in P =2x 2% + 2 % = 6 %

the amount of current in the coil

molecules are free to separate from each other and will take on the shape of the container. molecules move completely at random and high speeds. they vibrate in a fixed position and do not move around much. molecules expand to the edge of their container.