What potential difference is measured across a 19.9 Ω load resistor when it is connected across a battery of emf 2.46 V and internal resistance 0.561 Ω? Answer in units of V.

The potential difference measured is 2.388 V.

Explanation:

E = I(R + r)

E is the emf (electromotive force) of the battery = 2.46 V

R is the resistance of the resistor = 19.9 ohms

r is the internal resistance = 0.561 ohm

I (current) = E/(R + r) = 2.46/(19.9 + 0.561) = 2.46/20.461 = 0.12 A

V (potential difference) = IR = 0.12 × 19.9 = 2.388 V

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