The-resistance-R-of-an-unknown-resistor-is-found-by-measuring-the-potential-difference-V-across-the-resistor-and-the-current-I-through-it-and-using-the-equation-R-V-I-The-voltmeter-reading-has-a-3

Question Number 33272 by NECx last updated on 14/Apr/18

T h e r e s i s t a n c e R o f a n

u n k n o w n r e s i s t o r i s f o u n d b y

m e a s u r i n g t h e p o t e n t i a l

d i f f e r e n c e V a c r o s s t h e

r e s i s t o r a n d t h e c u r r e n t I t h r o u g h

i t a n d u s i n g t h e e q u a t i o n R = \frac{V}{I} .

T h e v o l t m e t e r r e a d i n g h a s a 3 %

u n c e r t a i n t y a n d t h e a m m e t e r

r e a d i n g h a s a 2 % u n c e r t a i n t y .

w h a t i s t h e u n c e r t a i n t y i n t h e

c a l c u l a t e d r e s i s t a n c e ?

Commented by NECx last updated on 14/Apr/18

p l e a s e h e l p

Answered by MJS last updated on 14/Apr/18

R = \frac{1.03 V}{. 98 I} \approx 1.05 \frac{V}{I}

R = \frac{. 97 V}{1.02 I} \approx . 95 \frac{V}{I}

so the uncertainty in R is 5 %

with both division and multiplication

the resulting uncertainty is the

sum of the single uncertainties

Commented by NECx last updated on 14/Apr/18

T h a n k s s o m u c h

Commented by NECx last updated on 14/Apr/18

T h i s w h a t I d i d t h o I d o n t k n o w

w h a t w e n t w r o n g

R = \frac{V}{I} \to \frac{d R}{R} = \frac{d V}{V} - \frac{d I}{I}

d R = R (\frac{d V}{V} - \frac{d I}{I})

d V = 3 % V d I = 2 % I

d R = R (3 % - 2 %)

d R = 1 % R

p l e a s e w h a t s w r o n g

Commented by MJS last updated on 14/Apr/18

{it}^{'} s not + 3 % and + 2 %,

{it}^{'} s \pm 3 % and \pm 2 % \Rightarrow

\Rightarrow d V = 3 % or d V = - 3 %

d I = 2 % or d I = - 2 %

with my method :

it could be \frac{1.03 V}{1.02 I} or \frac{. 97 V}{. 98 I} but

the maximum is higher, as I

showed above

Facebook Tweet Pin