Q.A resistor of 500 Ω and an inductance of 0.5 H are in series with an AC source which is given by V=1002sin(1000t). The power factor of the combination is (A) 21 (B) 31 (C) 0.5 (D) 0.6
Because both instantaneous voltage e and instantaneous current i constantly change in an AC circuit, the instantaneous power P=ei fluctuates continuously and is not, by itself, a physically meaningful description of energy consumption; what matters instead is the AVERAGE power over one complete cycle. Working this out separately for each basic element: a pure resistor (e and i in phase) gives Pav=21e0i0=ermsirms, the familiar 'apparent power' relation; a pure inductor or pure capacitor (each with a π/2 phase difference) gives EXACTLY ZERO average power, since the power expression reduces to a zero-mean sinusoid at twice the original frequency -- energy is merely exchanged cyclically with the source, never permanently consumed.
For the general case of a phase difference ϕ between voltage and current (as in a series LCR circuit), the average power is Pav=ermsirmscosϕ, where cosϕ -- the POWER FACTOR -- can also be read geometrically off the impedance triangle as R/Z. A power factor of 1 (purely resistive, or exactly at resonance) means all the apparent power is genuinely dissipated; a power factor of 0 (purely reactive, L or C alone) means the current, however large, dissipates no real power at all -- such a current is called WATTLESS or IDLE current. In practice, a low power factor forces a disproportionately larger current to deliver the same real power, causing much greater I2R transmission losses, which is why keeping the power factor close to 1 matters for real electrical systems.
[!TLDR] XL=ωL=1000×0.5=500Ω=R, so tanϕ=XL/R=1, ϕ=45∘, and the power factor is cos45∘=1/2. [!ANSWER] (A) 1/2
The source is V=1002sin(1000t), so ω=1000 rad/s. With L=0.5 H, the inductive reactance is XL=ωL=1000×0.5=500Ω, which happens to equal R=500Ω. There is no capacitor in this circuit, so it is a series R-L combination with impedance Z=R2+XL2=5002+5002=5002Ω. The power factor is cosϕ=ZR=5002500=21≈0.707. [!ANSWER] (A) 1/2≈0.707
Compute XL=ωL from the source's angular frequency, then use power factor =cosϕ=R/Z=R/R2+XL2 for this series R-L circuit.
Trying to include a capacitive reactance term when no capacitor is present in this circuit, or confusing power factor cosϕ with tanϕ=XL/R.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2025Set ANNUAL4 marks
Q.Obtain an expression for average power dissipated in a series LCR circuit.
›Reveal solutionSolution
The average power in an AC circuit is not simply VrmsIrms but includes the power factor cosφ accounting for the phase lag between current and voltage.
In a series LCR circuit driven by v=V0sinωt, the current is i=I0sin(ωt−φ), where φ is the phase angle between voltage and current, tanφ=(XL−XC)/R.
Instantaneous power:
p=vi=V0I0sinωtsin(ωt−φ)
Using sin(ωt−φ)=sinωtcosφ−cosωtsinφ:
p=V0I0[sin2ωtcosφ−sinωtcosωtsinφ]
Averaging over a full cycle: ⟨sin2ωt⟩=21 and ⟨sinωtcosωt⟩=0, so
where cosφ=R/Z is called the power factor of the circuit (Z=R2+(XL−XC)2 is the impedance). Only the component of current in phase with the voltage contributes to the average (real) power; the reactive (out-of-phase) component contributes zero average power.