Power dissipation limit in the load resistor

Steps:

• Maximum power P = I²R = 0.5 W, with R = 4 Ω, so I_max = √(0.5 / 4) = √0.125 ≈ 0.354 A.
• Total circuit resistance = external R + internal r = 4 Ω + 0.5 Ω = 4.5 Ω.
• Current I = E / 4.5 Ω, set I = I_max for limit, so E_max = 0.354 A × 4.5 Ω ≈ 1.59 V.
• Among options, 1.59 V is closest to 1.7 V.

Why C is correct:

• Calculated E_max ≈ 1.59 V matches 1.7 V as the nearest option per the power formula P = [E R / (R + r)]² / R.

Why the others are wrong:

• A: 1 V gives P ≈ 0.20 W, well below limit, not maximum E.
• B: 1.5 V gives P ≈ 0.44 W, below limit but underestimates maximum E.
• D: 2.9 V gives P ≈ 1.66 W, exceeds limit and overheats.

Final answer: C