Question

A battery of emf 12 V and internal resistance 3 \(\Omega\) is connected to an external resistor. If the current in the circuit is 0.6 A, the voltage across the external resistor will be

• A. 10.2 V
• B. 17.0 V
• C. 12.0 V
• D. 13.8 V

Hint:

Remember that the EMF (\(\mathcal{E}\)) is the total voltage the battery can provide, while the terminal voltage (\(V_T\)) is the actual voltage delivered to the external circuit. The difference, \(Ir\), is the 'lost volts' due to the battery's own internal resistance. The terminal voltage will always be less than the EMF when the battery is discharging (supplying current).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem deals with a simple DC circuit containing a real battery (which has both an electromotive force, EMF, and an internal resistance) and an external resistor. The voltage available to the external circuit (the terminal voltage) is the EMF minus the voltage drop across the internal resistance.

Step 2: Key Formula or Approach:
The terminal voltage \(V_T\) of a battery supplying a current \(I\) is given by: \[ V_T = \mathcal{E} - Ir \] where \(\mathcal{E}\) is the EMF of the battery and \(r\) is its internal resistance.
The voltage across the external resistor is equal to the terminal voltage of the battery.

Step 3: Detailed Explanation:
Given data:
EMF, \(\mathcal{E} = 12 \, \text{V}\).
Internal resistance, \(r = 3 \, \Omega\).
Current in the circuit, \(I = 0.6 \, \text{A}\).
Calculation:
The voltage across the external resistor is the terminal voltage \(V_T\). We can calculate this using the formula: \[ V_T = \mathcal{E} - Ir \] Substitute the given values: \[ V_T = 12 \, \text{V} - (0.6 \, \text{A} \times 3 \, \Omega) \] \[ V_T = 12 \, \text{V} - 1.8 \, \text{V} \] \[ V_T = 10.2 \, \text{V} \]

Step 4: Final Answer:
The voltage across the external resistor will be 10.2 V.