# Simple circuit with resistances

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.

• Are you using measured values of R1 and R2 to calculate the total resistance? Commented Oct 9, 2013 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) Commented Oct 9, 2013 at 0:54
• So yeah measured values :) Commented Oct 9, 2013 at 1:37

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