Q.The average power dissipated in a pure inductor is

Showing the 12 most recent of 21 on this concept.

1. CBSE 2025Set ANNUAL1 markMCQ
   Q.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 $P_{avg} = V_{rms} I_{rms}\cos\phi$. In a purely inductive circuit the current lags the applied voltage by exactly $90^\circ$, so $\cos\phi = \cos 90^\circ = 0$. Hence $P_{avg} = 0$: over a full cycle e …

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2. CBSE 2024Set A1 markMCQ
   Q.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

   $P_{avg} = V_{rms}\,I_{rms}\cos\phi,$

   where φ is the phase difference between the applied emf and the current. The factor $\cos\phi$ that determines what fraction of the apparent power $V_{rms}I_{rms}$ is actually consumed is called the power factor.

   …

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3. CBSE 2024Set A1 markMCQ
   Q.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 $P = V_{rms}I_{rms}\cos\phi$.

   • In a resistor, current and voltage are in phase (φ = 0), so cos φ = 1 and power $I_{rms}^2R$ is dissipated as heat. …

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4. CBSE 2024Set A1 mark
   Q.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 $P_{avg} = V_{rms}I_{rms}\cos\phi$, where $\cos\phi$ 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:

   $\cos\phi = \cos 0^\circ = 1$ …

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5. CBSE 2024Set ANNUAL1 markMCQ
   Q.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 …

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6. CBSE 2024Set ANNUAL1 mark
   Q.What is wattless current?

   ›Reveal solutionSolution

   The quadrature (reactive) current component, $I_{rms}\sin\phi$, that transfers no net energy.

   In an AC circuit with a phase difference $\phi$ between voltage and current, the average power is $P = V_{rms}I_{rms}\cos\phi$. The current can be resolved into two components: one in phase with the voltage, $I_{rms}\cos\phi$ (which does work / consumes power), and one $90°$ out of phase with the voltage, $I_{rms}\sin\phi$ (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 …

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7. CBSE 2023Set MODEL1 markMCQ
   Q.In an LCR circuit, the power factor becomes 0 (zero) when:

   (a) $R = 0$

   (b) $\omega L = \omega C$

   (c) $\omega L = \dfrac{1}{\omega C}$

   (d) $\left(\omega L - \dfrac{1}{\omega C}\right) = R$

   ›Reveal solutionSolution

   Power factor is zero only for a purely reactive (resistance-less) AC circuit.

   In an LCR circuit, the power factor is $\cos\phi = \dfrac{R}{Z}$, where $Z=\sqrt{R^2+(\omega L-\tfrac{1}{\omega C})^2}$. The power factor becomes zero (i.e. $\phi = 90^\circ$, purely reactive circuit, average power consumed is zero) only when $R=0$. …

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8. CBSE 2023Set B1 mark
   Q.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 $\pi/2$. The instantaneous power is $p = vi = v_0 i_0 \sin(\omega t)\sin(\omega t - \pi/2) = -v_0 i_0 \sin(\omega t)\cos(\omega t)$, which when averaged over a complete cycle integrates to zero because $\langle \sin\omega t\cos\omega t\rangle = 0$ over a full period. Physically, the inductor stores energy in its magnetic field during one quarter cycle …

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9. CBSE 2022Set I1 markMCQ
   Q.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:

   $P_{av} = E_{rms}\,I_{rms}\cos\varphi$

   …

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10. CBSE 2022Set I1 markMCQ
    Q.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:

    $\cos\varphi = \frac{R}{Z}$

    …

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11. CBSE 2022Set ANNUAL1 markMCQ
    Q.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.

    …

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12. CBSE 2021Set A1 markMCQ
    Q.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:

    $\text{power factor} = \frac{\text{True power}}{\text{Apparent power}} = \cos\varphi$

    …

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