Measuring e.m.f. and Internal Resistance - Simultaneous Equation

Example 1
When a 1Ω resistor is connected to the terminal of a cell, the current that flow through it is 8A. When the resistor is replaced by another resistor with resistance 4Ω, the current becomes 2⅔A. Find the
a. internal resistance of the cell
b. e.m.f. of the cell

Answer:
Experiment 1
R₁ = 1Ω
I₁ = 8A
$$\begin{array}{l}E=IR+Ir\\ E=(8)(1)+(8)r\\ E-8r=8\end{array}$$
Experiment 2
R₂ = 4Ω
I₂ = 2⅔A
$$\begin{array}{l}E=IR+Ir\\ E=(2\frac{2}{3})(4)+(2\frac{2}{3})r\\ 3E-8r=32\end{array}$$
Solve the simultaneous equation
E = 12V, r = 0.5Ω

Example 2
The diagram on the left shows that the terminal potential difference of a batteries is 1.2V when a 4 Ω resistor is connected to it. The terminal potential become 1.45V when the resistor is replaced by another resistor of resistance 29Ω
Find the
a. internal resistance, r
b. e.m.f. of the batteries.
Answer:
Experiment 1
V₁ = 1.2V
R₁ = 4Ω
$$\begin{array}{l}I=\frac{V}{R}\\ I=\frac{1.2}{4}=0.3A\end{array}$$
$$\begin{array}{l}E=V+Ir\\ E=1.2+0.3r\\ E-0.3r=1.2\end{array}$$
Experiment 2
V₂ = 1.45V
R₂ = 29
$$\begin{array}{l}I=\frac{V}{R}\\ I=\frac{1.45}{29}=0.05A\end{array}$$
$$\begin{array}{l}E=V+Ir\\ E=1.45+0.05r\\ E-0.05r=1.45\end{array}$$
Solve the simultaneous equation
E = 1.5V, r = 1Ω