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

Question

3 years ago

8

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.

Physics

2 answers:

lesya692 [45] · Answer 1 · 2022-05-18T07:07:55+03:00

lesya692 [45]3 years ago

8 0

Here is a hint for you
P=VI=IR*I=I^2R

Ostrovityanka [42] · Answer 2 · 2022-05-18T08:53:50+03:00

Ostrovityanka [42]3 years ago

3 0

Answer:

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 %