A battery of e.m.f. $12$ V and internal resistance $1\ \Omega$ is connected to a $5\ \Omega$ resistor. Find the terminal potential difference of the battery.
A$2.4$ V
B$12$ V
C$10$ VCORRECT
D$2$ V
JUPEBPhysicsCurrent Electricity
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Why the answer is C, and why the others tempt you.
Current $I = \dfrac{E}{R + r} = \dfrac{12}{5 + 1} = 2$ A. Terminal p.d. $= IR = 2 \times 5 = 10$ V, or equivalently $E - Ir = 12 - 2 = 10$ V. The 2 V is the 'lost volts' across the internal resistance.
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