Q.In the given circuit, calculate
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Average Power Absorption
Average Power Absorption – From Intuition to Precision
Think of pushing a child on a swing. You don't push constantly — you push only when the swing is moving away from you, and you time your push to add energy each time. Some pushes land perfectly, others might be slightly off. Over several minutes, what matters is not the force at any single instant, but the net energy you transferred averaged over time.
That's the core idea behind average power absorption: how much energy, on average, is being delivered per unit time to a device or system, even when the instantaneous power fluctuates wildly.
The Intuitive Picture
Consider a light bulb connected to household AC supply. The voltage oscillates 50 times per second (in India). At the peak of the voltage cycle, the bulb glows brightest; when voltage crosses zero, the bulb goes dark for an instant. But you don't see flickering — your eyes average out the rapid changes. What you perceive as "brightness" corresponds to the average power the bulb absorbs.
Similarly, when you charge a phone battery, the power drawn isn't constant — it's high when the battery is low, then tapers off. The "charging speed" you care about is the average power over the charging session.
The Precise Definition
Pavg=T1∫0Tp(t)dt
Where:
- p(t) is the instantaneous power at time t (in watts)
- T is the time period over which we average (in seconds)
For a resistor with a sinusoidal voltage v(t)=Vmsin(ωt) and current i(t)=Imsin(ωt) (since they're in phase), the instantaneous power is:
p(t)=v(t)⋅i(t)=VmImsin2(ωt)
This is always positive (since sin2 is never negative) but it oscillates between 0 and VmIm. The average over one complete cycle gives:
Pavg=2VmIm
Why This Matters for Exams
The most common mistake students make is confusing peak power with average power. A 100 W bulb doesn't draw 100 W at every instant — it draws about 200 W at the voltage peak and 0 W at the zero crossing. The 100 W rating is the average power it's designed to dissipate safely.
Never use P=VI directly with AC peak values unless you divide by 2 (for sinusoidal waveforms). The correct formula for average power in a resistor is Pavg=2VmIm=VrmsIrms, where Vrms=Vm/2.
The General Case (Phase Differences)
When voltage and current are not in phase — as in circuits with inductors or capacitors — the instantaneous power can become negative during parts of the cycle (energy flows back to the source). The average power then becomes:
Pavg=VrmsIrmscosϕ …
Why this formula?
Average Power Absorption: Why the Formula Holds
Let's build this from first principles — understanding why average power is what it is, not just memorising the formula.
1. Instantaneous Power — The Starting Point
For any circuit element, instantaneous power is always:
p(t)=v(t)⋅i(t)
This is the fundamental definition: power at an instant is voltage times current at that same instant.
2. Why We Need an Average
In AC circuits, both v(t) and i(t) vary sinusoidally with time. So p(t) also varies — often at twice the frequency of the original signals.
- Instantaneous power oscillates between zero and a peak value.
- What matters for real energy consumption is the average over a complete cycle.
Hence, we define:
Pavg=T1∫0Tp(t)dt
where T is the time period of the AC waveform.
3. The Key Derivation (Step-by-Step)
Step 1: Write the sinusoidal forms
Let:
- v(t)=Vmcos(ωt+θv)
- i(t)=Imcos(ωt+θi)
Here θv and θi are phase angles. The phase difference is:
ϕ=θv−θi
Step 2: Instantaneous power
p(t)=VmImcos(ωt+θv)cos(ωt+θi)
Use the trigonometric identity:
cosAcosB=21[cos(A−B)+cos(A+B)]
So:
p(t)=2VmIm[cos(θv−θi)+cos(2ωt+θv+θi)]
Step 3: Average over one cycle
The average of cos(2ωt+constant) over a full cycle is zero — because it's a sinusoid symmetric about zero.
Only the constant term survives:
Pavg=2VmImcos(ϕ)
4. The Standard Form Using RMS Values
Recall:
- Vrms=2Vm
- Irms=2Im
Therefore:
2VmIm=VrmsIrms
So the final formula is:
Pavg=VrmsIrmscosϕ
5. What cosϕ Really Means
- ϕ is the phase difference between voltage and current.
- cosϕ is called the power factor.
- Why it appears: Only the component of current in phase with voltage contributes to average power. The quadrature (90° out-of-phase) component averages to zero.
| ϕ | cosϕ | Interpretation | …
E = √(V_R² + (V_L − V_C)²) = 50 V; power factor cosφ = V_R/E = 0.8.
In a series LCR a.c. circuit the voltages across R, L and C are not in phase: V_R is in phase with the current, V_L leads by 90°, and V_C lags by 90°. So they add as phasors, not arithmetically.
(i) Source emf: The applied emf equals the phasor sum:
E = √( V_R² + (V_L − V_C)² )
E = √( 40² + (50 − 20)² )
E = √( 1600 + 900 )
E = √2500 = 50 V.
…
Showing the 12 most recent of 21 on this concept.
- CBSE 2025Set ANNUAL1 markMCQQ.The average power dissipated in a pure inductor is(i) VI^2(ii) zero(iii) (1/2)VI(iv) VI^2/4
›Reveal solutionSolution
A pure inductor has phase angle 90 degrees, so average power = V I cos(90) = 0.
Average power in an AC circuit is Pavg=VrmsIrmscosϕ. In a purely inductive circuit the current lags the applied voltage by exactly 90∘, so cosϕ=cos90∘=0. Hence Pavg=0: over a full cycle e …
- CBSE 2024Set A1 markMCQQ.If the phase difference between alternating current and e.m.f. is φ, then the value of power factor is (A) cos φ (B) cos^2 φ (C) sin φ (D) tan φ
›Reveal solutionSolution
Power factor = cos φ.
The average power dissipated in an AC circuit is
Pavg=VrmsIrmscosϕ,
where φ is the phase difference between the applied emf and the current. The factor cosϕ that determines what fraction of the apparent power VrmsIrms is actually consumed is called the power factor.
…
- CBSE 2024Set A1 markMCQQ.In AC circuit, power is lost in only (A) resistance (B) inductance (C) capacitance (D) all of these
›Reveal solutionSolution
Only resistance dissipates (real) power in AC; ideal L and C do not.
In an AC circuit the average power is P=VrmsIrmscosϕ.
- In a resistor, current and voltage are in phase (φ = 0), so cos φ = 1 and power Irms2R is dissipated as heat. …
- CBSE 2024Set A1 markQ.Write the value of power factor for a pure resistive circuit.
›Reveal solutionSolution
In a pure resistor, V and I are in phase, so power factor cos φ = cos 0° = 1.
The average power dissipated in an AC circuit is Pavg=VrmsIrmscosϕ, where cosϕ is called the power factor and φ is the phase difference between the current and the voltage.
In a purely resistive circuit, the current is always in phase with the applied voltage (φ = 0°), because a resistor offers no reactance. Therefore:
cosϕ=cos0∘=1 …
- CBSE 2024Set ANNUAL1 markMCQQ.For an alternating current i = I0 sinωt passing through a resistor R, how much is the average power loss due to Joule heating ?(a) I0²R(b) (1/2) I0²R(c) 4 I0²R(d) 2 I0²R
›Reveal solutionSolution
Average power in a resistor over one AC cycle is Irms²R, and Irms = I0/√2 for a sinusoidal current.
For i = I0 sin ωt through a resistor R, the instantaneous power is p = i²R = I0²R sin²ωt.
Averaging sin²ωt over a full cycle gives 1/2, so:
Pavg = I0²R × (1/2) = (1/2) I0²R …
- CBSE 2024Set ANNUAL1 markQ.What is wattless current?
›Reveal solutionSolution
The quadrature (reactive) current component, Irmssinϕ, that transfers no net energy.
In an AC circuit with a phase difference ϕ between voltage and current, the average power is P=VrmsIrmscosϕ. The current can be resolved into two components: one in phase with the voltage, Irmscosϕ (which does work / consumes power), and one 90° out of phase with the voltage, Irmssinϕ (the quadrature component). This second component does no net work over a complete cycle — it is called the wattless current (or idle current). It occurs purely in ideal (resistance-free) inductive or capacitive circuits, w …
- CBSE 2023Set MODEL1 markMCQQ.In an LCR circuit, the power factor becomes 0 (zero) when:(a) R=0(b) ωL=ωC(c) ωL=ωC1(d) (ωL−ωC1)=R
›Reveal solutionSolution
Power factor is zero only for a purely reactive (resistance-less) AC circuit.
In an LCR circuit, the power factor is cosϕ=ZR, where Z=R2+(ωL−ωC1)2. The power factor becomes zero (i.e. ϕ=90∘, purely reactive circuit, average power consumed is zero) only when R=0. …
- CBSE 2023Set B1 markQ.Write True or False: The average power supplied to an inductor over one complete cycle is zero.
›Reveal solutionSolution
Because current and voltage in a pure inductor are 90° out of phase, the average power delivered to it over one full cycle works out to zero.
For a pure inductor in an AC circuit, current lags voltage by π/2. The instantaneous power is p=vi=v0i0sin(ωt)sin(ωt−π/2)=−v0i0sin(ωt)cos(ωt), which when averaged over a complete cycle integrates to zero because ⟨sinωtcosωt⟩=0 over a full period. Physically, the inductor stores energy in its magnetic field during one quarter cycle …
- CBSE 2022Set I1 markMCQQ.The expression for real average power in an alternating current circuit is (A) P_av = E_rms cos φ (B) P_av = I_rms cos φ (C) P_av = E_rms I_rms sin φ (D) P_av = E_rms I_rms cos φ
›Reveal solutionSolution
Average power in AC = E_rms × I_rms × cos φ, where cos φ is the power factor.
In an AC circuit the instantaneous voltage and current are generally out of phase by an angle φ. Averaging the instantaneous power e·i over a full cycle gives the real (active) power:
Pav=ErmsIrmscosφ
…
- CBSE 2022Set I1 markMCQQ.Power factor is equal to (A) R/Z (B) Z/R (C) R.Z (D) W.L
›Reveal solutionSolution
Power factor cos φ = R/Z.
In a series AC circuit the impedance triangle has resistance R along the base, net reactance (X_L − X_C) as the perpendicular, and impedance Z as the hypotenuse. The phase angle φ between voltage and current satisfies:
cosφ=ZR
…
- CBSE 2022Set ANNUAL1 markMCQQ.In purely inductive or capacitive circuit, power factor (cos φ) is ____.(a) 0(b) 1(c) -1
›Reveal solutionSolution
In a purely inductive or purely capacitive AC circuit, current and voltage are 90° out of phase, so the power factor is zero.
…
- CBSE 2021Set A1 markMCQQ.Which of the following relations is correct for power factor? (A) Power factor = True average power × Apparent average power (B) Power factor = Apparent average power / True average power (C) Power factor = True average power / Apparent average power (D) Power factor = ½ [ True average power × Apparent average power ]
›Reveal solutionSolution
Power factor = true (real) power ÷ apparent power = cos φ.
In an ac circuit the apparent power is V_rms·I_rms, while the true (real) power actually consumed is V_rms·I_rms·cos φ, where φ is the phase angle between voltage and current. The power factor is the ratio of these:
power factor=Apparent powerTrue power=cosφ
…
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