The terminals of a battery of emf 24 0 V and in... - JAMB Physics 2023 Question

The terminal potential difference (\(V\)) of a battery connected to an external resistor can be calculated using the formula:

\[V = \varepsilon - Ir\]

where:
- \(\varepsilon\) is the electromotive force (emf) of the battery,
- \(I\) is the current flowing through the circuit,
- \(r\) is the internal resistance of the battery.

Given that \(\varepsilon = 24.0 \, \text{V}\), \(r = 1.0 \, \Omega\), and the external resistor is \(5.0 \, \Omega\), we need to find the current (\(I\)) first using Ohm's Law:

\[I = \frac{\varepsilon}{R_T}\]

where \(R_T\) is the total resistance, which is the sum of the external resistor and the internal resistance:

\[R_T = R + r\]

\[R_T = 5.0 + 1.0 = 6.0 \, \Omega\]

Now, calculate \(I\):

\[I = \frac{24.0}{6.0} = 4.0 \, \text{A}\]

Now, substitute \(I\) and the given values into the first formula for \(V\):

\[V = 24.0 - (4.0 \times 1.0) = 20.0 \, \text{V}\]

Therefore, the terminal potential difference is \(20.0 \, \text{V}\).

So, the correct option is: 20.0V