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The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 Ω, when connected through an external resistance of 4 Ω as shown in the figure is:
V = E × R / (R + r) = 10 × 4 / (4 + 1) = 40 / 5 = 8 V!
According to Ohm’s law applied to the complete closed circuit consisting of the real battery and the external resistor, the current is:
During discharge through an external load, the terminal voltage V across the battery terminals is reduced from its open-circuit EMF by the internal voltage drop Ir:
The terminal voltage must also identically equal the potential difference across the external 4 Ω resistor:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Exact terminal voltage: 10 V - 2 V = 8 V.
This is the open-circuit EMF, ignoring internal resistance drop.
Incorrect; confuses resistance value (4 Ω) with voltage.
Calculation mistake; occurs if internal resistance drop is miscalculated as 4 V.
Candidates often select 10 V, assuming the battery always delivers its rated EMF regardless of the load.
"The terminal voltage across a cell in discharging mode is given by V = E - Ir, where Ir represents the internal ohmic potential drop."
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