Question

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.

Answer 1

Hint: P=VI=IR*I=I^2R

Answer 2

now, Find percentage uncertanity in current

Answer 3

can u?

Answer 4

trying ...

Answer 5

Ok. let me help: (0.05/2.50 * 100)%=

Answer 6

thats 2%..... right?

Answer 7

got it ..

Answer 8

Now, Total percentage error in the power calculated = 2* 2% + 2%=6%

Answer 9

right?

Answer 10

yes ..i did it ... :)

Answer 11

Great

Answer 12

Did u understand? what I did

Answer 13

yes multiplied the uncertainty which we got by solving with the uncertainty of value obtained ...right?

Answer 14

Its actually because

Answer 15

So, percentage uncertainity in Power= 2 times the percentage uncertainity in Current + percentage uncertainity in resistance

Answer 16

got it?

Answer 17

okay

Answer 18

Do u have any other problem?

Answer 19

yes wait

Answer 20

First close this question