Circuit Exercise-6: Solution

Assume a 15-Ω light bulb powered by two 1.5-V batteries in series.
Question:
Calculate the power dissipated

Solution:
ε = ε₁ + ε₂ = 1.5 V + 1.5 V = 3 V
R = 15Ω
ε = IR ⇒ I = ε / R = 3 / 15 = 0.2 A
P = εI = I²R = (0.2 A)² × 15 Ω= 0.6 W

Now, let's double check:
P = εI = (3 V) × 0.2 A = 0.6 W

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Date of last modification: 2022