Power factor and real power in a series RL circuit (concept check): At a fixed RMS source voltage, does a power factor of 0.5 result in less average real power (heat in R) than a power factor of 0.6?

Introduction / Context:
Power factor (pf) describes the fraction of apparent power that becomes real power (heat and useful work). In a series RL circuit supplied by a fixed RMS voltage, the higher the power factor, the larger the real power delivered to the resistor for a given apparent power. This item compares pf = 0.5 and pf = 0.6 under constant supply voltage.

Given Data / Assumptions:

• Sinusoidal steady state with fixed RMS source voltage V.
• Series RL load; real power dissipates only in R.
• Power factor pf = cos(theta) with theta = arctan(XL / R).

Concept / Approach:
Real power P is P = V * I * pf. At fixed V, apparent power S = V * I, while pf determines how much of S becomes P. For the same network at different operating points (or different L/R combinations) resulting in pf = 0.5 vs 0.6, the case with pf = 0.6 converts a larger portion of the apparent power into real power, hence more heat in R and a higher meter reading on a wattmeter.

Step-by-Step Solution:

Verification / Alternative check:
With normalized S = 1 VA, pf = 0.5 gives P = 0.5 W; pf = 0.6 gives P = 0.6 W. The trend is general: increasing pf (approaching unity) increases real power for a given apparent power or given supply voltage with similar loading.

Why Other Options Are Wrong:
“More power at 0.5” contradicts P = S * pf. “Same power” is only true if other variables change to offset pf, which is not implied here. “Need L” is unnecessary for the qualitative comparison.

Common Pitfalls:
Confusing fixed current with fixed voltage. At fixed current, P = I^2 * R and pf discussion differs; the question specifies fixed voltage supply context.

Final Answer:
Applies (pf = 0.5 yields less real power than pf = 0.6)