For a battery of emf ε and internal resistance r driving current I through external resistance R, the terminal voltage V equals
A{'text': 'ε + I r', 'label': 'A'}
B{'text': 'ε × I r', 'label': 'B'}
C{'text': 'ε − I r', 'label': 'C'}
D{'text': 'ε / (I r)', 'label': 'D'}
Answer & Solution
Correct answer: C. {'text': 'ε − I r', 'label': 'C'}
1. The battery emf drops across both external R and internal r.
2. Terminal voltage V is the potential difference across the external resistance.
3. V = ε − I r, where I r is the internal potential drop.
4. When the battery is being discharged, V < ε; on open circuit V = ε.
_Source: NCERT Class 12 Physics, Ch 3 "Current Electricity", §3 EMF and internal resistance_
Related questions
The reactance of a capacitor $C$ at frequency $f$ in an a.c. circuit is:A bird perched on a single high voltage wire is unharmed because:The two conservation laws embodied in Kirchhoff's rules are conservation of:The resistance offered by the cell itself is called:Which device detects the current in a Wheatstone bridge?In a Wheatstone bridge, the source is connected across the:The Wheatstone bridge is made up of how many resistors?At a junction, the current entering equals the current: