Step 1: Concept Overview:
This problem concerns a basic DC circuit featuring a real battery (characterized by electromotive force, EMF, and internal resistance) and an external resistor. The voltage available to the external circuit, known as the terminal voltage, is derived by subtracting the voltage drop across the internal resistance from the EMF.
Step 2: Applicable Formula:
The terminal voltage \(V_T\) of a battery delivering a current \(I\) is defined by the equation:\[ V_T = \mathcal{E} - Ir \]Here, \(\mathcal{E}\) represents the battery's EMF, and \(r\) denotes its internal resistance.
Consequently, the voltage across the external resistor is equivalent to the battery's terminal voltage.
Step 3: Detailed Calculation:
Provided Data:
EMF, \(\mathcal{E} = 12 \, \text{V}\).
Internal resistance, \(r = 3 \, \Omega\).
Circuit current, \(I = 0.6 \, \text{A}\).
Computation:
The voltage across the external resistor is identical to the terminal voltage \(V_T\). This can be calculated using the formula:\[ V_T = \mathcal{E} - Ir \]Substituting the given values yields:\[ V_T = 12 \, \text{V} - (0.6 \, \text{A} \times 3 \, \Omega) \]\[ V_T = 12 \, \text{V} - 1.8 \, \text{V} \]\[ V_T = 10.2 \, \text{V} \]
Step 4: Conclusive Result:
The voltage measured across the external resistor is 10.2 V.