# Measurement of Electromotive Force and Internal Resistance

test page

Connect the circuit in Figure 0. According to Ohmâ€™s law:

$$

\varepsilon=I(R+R_A+r)

$$

where $I$ is the current through the conductor(milliammeter reading), $R$ is the resistance value of the resistor box, $R_A$ is the size of the internal resistance of the milliammeter, $r$ is the size of the internal resistance of the power supply and $\varepsilon$ is the electromotive force of the power supply.

So there is:

$$

\frac{1}{I}=\frac{R}{\varepsilon}+\frac{R_g+r}{\varepsilon}

$$

Because $\displaystyle{\frac{1}{I}}$ and $R$ are linear, we can change the resistance value of the resistor box to read out multiple sets of $\displaystyle{\frac{1}{I}}$ and $R$ to make $\displaystyle{\frac{1}{I}} \sim R$ image and get the slope $k_1$, so

$$

\varepsilon=\frac{1}{k_1}

$$

Then connect the circuit in Figure 1, and use Ohm's law again: $$ \frac{\varepsilon}{R+R_v+r}=\frac{U}{R_v}=I $$ where $R_V$ is the size of the internal resistance of the voltmeter and $U$ is the reading of voltmeter. So, $$ \frac{\varepsilon}{U}=1+\frac{R}{R_V}+\frac{r}{R_V} $$ From the above equation, $\displaystyle{\frac{\varepsilon}{U}}$ and $R$ are also linearly related, so we can make an $\displaystyle{\frac{\varepsilon}{U}} \sim R$ plot and find the slope $k_2$ and intercept $b_2$ after obtaining several sets of data. Then, $$ \begin{cases} k_2=\displaystyle{\frac{1}{R_V}}, \\\\ b_2=1+\displaystyle{\frac{r}{R_V}} \end{cases}\Rightarrow r=\displaystyle{\frac{b_2-1}{R_2}} $$