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A lamp consumes only 50% of maximum power in an A.C. circuit. What is the phase difference between the applied voltage and the circuit current?

Asked in GUJCET 2018 · Power factor

• (1) π/4
• (2) π/3
• (3) π/6
• (4) π/2

Step-by-step solution

Why the other options are wrong

• (1) π/4 needs cos φ=1/(√2), i.e. power proportional to cos²φ. That happens only if the current also falls as the phase grows (a series circuit at its half-power points). Here the comparison is at the same Vᵣₘₛ and Iᵣₘₛ, so the factor is cos φ alone.
• (3) π/6 comes from setting sin φ=1/2. The power factor is cos φ, not sin φ.
• (4) At π/2 the power factor is zero and the lamp would take no power at all (a wattless current), not half.

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