alleryA 12V battery (internal r = 0.5 Ω) drives current through 5.5 Ω external load. Find power dissipated in EXTERNAL resistor:
A11 W
B22 W
C24 W
D8 W
Answer & Solution
Correct answer: B. 22 W
Total R = 5.5 + 0.5 = 6 Ω.
I = 12/6 = 2 A. Power in external = I² × R = 4 × 5.5 = 22 W.
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