# Measurement of Electromotive Force and Internal Resistance

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Figure 0

Connect the circuit in Figure 0. According to Ohm’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
$$\varepsilon=\frac{1}{k_1}$$

Figure 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, $$\begin{cases} k_2=\displaystyle{\frac{1}{R_V}}, \\\\ b_2=1+\displaystyle{\frac{r}{R_V}} \end{cases}\Rightarrow r=\displaystyle{\frac{b_2-1}{R_2}}$$
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