Answer:
EMF = 11.35 V
R = 0.031Ω
Explanation:
To find the battery's EMF and the internal resistance we need to use Ohm's law:
[tex] V = IR [/tex]
Where:
V: is the voltage
I: is the current
R is the resistance
We have:
The current through the battery is 64.2 A and the potential difference across the battery terminals is 9.36 V:
[tex] \epsilon = V + IR [/tex]
[tex] \epsilon = 9.36 V + 64.2A*R [/tex] (1)
When only the car's lights are used, the current through the battery is 1.96 A and the terminal potential difference is 11.3 V:
[tex] \epsilon = 11.3 V + 1.96A*R [/tex] (2)
By solving equation (1) and (2) for R we have:
[tex] 9.36 V + 64.2A*R = 11.3 V + 1.96A*R [/tex]
[tex]R = \frac{1.94 V}{62.24A} = 0.031 \Omega[/tex]
Hence, the internal resistance is 0.031 Ω.
Now, by entering R into equation (1) we can find the battery's EMF:
[tex]\epsilon = 9.36 V + 64.2A*0.031 \Omega[/tex]
[tex] \epsilon = 11.35 V [/tex]
Therefore, the battery's EMF is 11.35 V.
I hope it helps you!
