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mrs_skeptik [129]
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]3 years ago
8 0
Here is a hint for you
P=VI=IR*I=I^2R
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 %

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