Answer:
5 ohms
Explanation:
Given:
EMF of the ideal battery (E) = 60 V
Voltage across the terminals of the battery (V) = 40 V
Current across the terminals (I) = 4 A
Let the internal resistance be 'r'.
Now, we know that, the voltage drop in the battery is given as:
Therefore, the voltage across the terminals of the battery is given as:
![V= E-V_d\\\\V=E-Ir](https://tex.z-dn.net/?f=V%3D%20E-V_d%5C%5C%5C%5CV%3DE-Ir)
Now, rewriting in terms of 'r', we get:
![Ir=E-V\\\\r=\frac{E-V}{I}](https://tex.z-dn.net/?f=Ir%3DE-V%5C%5C%5C%5Cr%3D%5Cfrac%7BE-V%7D%7BI%7D)
Plug in the given values and solve for 'r'. This gives,
![r=\frac{60-40}{4}\\\\r=\frac{20}{4}\\\\r=5\ ohms](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B60-40%7D%7B4%7D%5C%5C%5C%5Cr%3D%5Cfrac%7B20%7D%7B4%7D%5C%5C%5C%5Cr%3D5%5C%20ohms)
Therefore, the internal resistance of the battery is 5 ohms.