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A 12 V battery with internal resistance 2 Ω is connected across a 4 Ω resistor. The TERMINAL voltage of the battery is
A4 V
B12 V
C8 V
D6 V
Answer & Solution
Correct answer: C. 8 V
I = 12/(4+2) = 2 A. Terminal V = ε − Ir = 12 − 2·2 = 8 V. Equivalently V = IR_ext = 2·4 = 8 V.
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