# 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

Experiment 1
R1 = 1Ω
I1 = 8A
$E=IR+Ir E=(8)(1)+(8)r E−8r=8$
Experiment 2
R2 = 4Ω
I2 = 2⅔A
$E=IR+Ir E=(2 2 3 )(4)+(2 2 3 )r 3E−8r=32$
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.
$I= V R I= 1.2 4 =0.3A$
$E=V+Ir E=1.2+0.3r E−0.3r=1.2$
$I= V R I= 1.45 29 =0.05A$
$E=V+Ir E=1.45+0.05r E−0.05r=1.45$