Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have a question about a circuit that I made in class!

So, we had a circuit with 2 resistors, one of $R_1 = 330 \Omega$ and one of $R_2 = 470 \Omega$. The tension applied to the circuit should be $10V$ but we applied $9.99V(V_f)$. When measuring the resistances with a multimeter the value for the $R_1$ was $328 \Omega$ and for the $R_2$ was $470 \Omega$. The value for both $R_1$ and $R_2$ in series was $R_1 + R_2 = 799 \Omega$.

The tension measured in $R_1$ was $4.02V(V_{r1})$ and in $R_2$ was $5.74V(V_{r2})$

The intensity of the current ($I$) was $12.34 mA$

With that said, we were asked to calculate $R_1$, $R_2$ and $R_1+R_2$ theoretically using the values of $V_{r1}$, $V_{r2}$ and $V_f$.

I used Ohm Law: $R = \frac VI$

$R_1 = 326 \Omega$ and $R_2 = 465 \Omega$

My question is: should be $R_1+R_2$ be calculate using $V_f (9.99/0.01234)$ as it was asked or calculating an equivalent resistance? By the first method the result is $810$ and be the second is $791$. The ammeter's resistance is $11 \Omega$ at the used scale $(20mA)$.

I'm asking cause I found it weird that $810$ is bigger than $799$(experimental value) given that the experimental should be bigger because of ammeter's resistance.

Sorry if the question is dumb, but I'm not a Physics student though I attend Physics classes.

share|improve this question
    
Are you using measured values of R1 and R2 to calculate the total resistance? –  mcodesmart Oct 9 '13 at 0:51
    
You mean the equivalent resistance I asked about? I used the values calculated with Ohm Law which is: R1 = 4,02 / (12,34 x 10^-3) and R2 = 5,74 / (12,34 x 10^-3) –  ipg24 Oct 9 '13 at 0:54
    
So yeah measured values :) –  ipg24 Oct 9 '13 at 1:37

1 Answer 1

up vote 0 down vote accepted

If you compute the relative error in that measurement:

$\frac{| R_{exp}-R_{theory}|}{/R_{theory}}=\frac{|810-799| \Omega}{799 \Omega}=1.3\%$ error

Resistor have an inherent amount of error in them, because of how they are made, or environmental conditions at the time of use. Assuming you were using pretty common components each of your resistor most likely had a 15-20\% built in error. (the error rating should be the final band on the resistor). Clearly your measured value and the theoretical value are in accord.

Also, when computing measured value for electronics measurements, you want to measure the resistance of each component independent of the circuit.

share|improve this answer
    
Ohh I did measure them independent of the circuit, the values I presented for the resistors (R1 = 328 and R2 = 470) were measured before putting them on the circuit :) The last band presents a 5% error rating which gives R1 a 16,5 error. But the error in the measure is 1. Its inside the boundaries specified by the "producer". –  ipg24 Oct 9 '13 at 1:58
    
Thanks for your answer, just one thing I don't understand if R = R1 + R2 + R.ampere, why isn't the experimental value of R1+R2 which includes R.ampere (799) less than the theorical value which is supposed not to contain R.ampere, independant of the error ratings? –  ipg24 Oct 9 '13 at 2:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.