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Wattless current, AC generator and transformer: NEET MCQs with answers

75 questions on this topic of Electromagnetic Induction and Alternating Currents (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. An ideal transformer has 1000 primary turns and 4000 secondary turns. If 220 V AC is applied to the primary, the secondary voltage is:

  1. 55 V
  2. 880 V
  3. 220 V
  4. 8.8 × 10⁵ V
Show answer and explanation

Answer: (B) 880 V

Vs/Vp = Ns/Np, so Vs = 220 × 4000/1000 = 880 V.

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2. A rectangular coil of an AC generator has N turns, area A, and rotates at constant angular frequency ω about an axis perpendicular to a uniform magnetic field B. When the plane of the coil makes an angle of 30° with the magnetic field lines, the magnitude of the induced emf is:

  1. N B A ω / 2
  2. (√3 / 2) N B A ω
  3. N B A ω
  4. (1 / √2) N B A ω
Show answer and explanation

Answer: (B) (√3 / 2) N B A ω

Magnetic flux is Φ = B A cos θ, where θ is the angle between the normal to the coil and the field B. If the plane makes an angle of 30° with the field, the normal makes an angle θ = 90° - 30° = 60° with the field, or in terms of time, θ = ωt. The instantaneous emf is ε = N B A ω sin(ωt) = N B A ω sin(60°) = (√3 / 2) N B A ω.

hard

3. A step-up transformer operates on a 220 V line and delivers a current of 2 A to an external load at 2200 V. If the primary winding has a resistance of 0.3 Ω and the secondary winding has a resistance of 15 Ω, what is the total copper loss in the windings?

  1. 120 W
  2. 180 W
  3. 60 W
  4. 240 W
Show answer and explanation

Answer: (B) 180 W

Assuming negligible power loss for turns ratio calculation, V_s / V_p = 2200 / 220 = 10. Thus, primary current is I_p ≈ I_s × 10 = 2 × 10 = 20 A. Copper loss in primary = (I_p)² R_p = (20)² × 0.3 = 400 × 0.3 = 120 W. Copper loss in secondary = (I_s)² R_s = (2)² × 15 = 4 × 15 = 60 W. Total copper loss = 120 W + 60 W = 180 W.

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4. A step-up transformer connected to a 250 V AC line has an efficiency of 80%. It delivers power to a load at 2000 V. If the current flowing in the secondary circuit is 0.5 A, the current drawn by the primary winding is:

  1. 5.0 A
  2. 4.0 A
  3. 2.5 A
  4. 6.25 A
Show answer and explanation

Answer: (A) 5.0 A

Output power is P_out = V_s × I_s = 2000 V × 0.5 A = 1000 W. With an efficiency η = 80%, input power is P_in = P_out / η = 1000 / 0.8 = 1250 W. Since P_in = V_p × I_p, we have 250 × I_p = 1250, giving I_p = 5.0 A.

easy

5. In an AC circuit containing an ideal inductor of inductance L connected to an alternating voltage source V = V₀ sin(ωt), which of the following statements is correct regarding the current?

  1. The current is completely wattless as the phase difference between voltage and current is π/2.
  2. The average power consumed over a full cycle is V₀I₀/2.
  3. The entire current is wattful because voltage and current are in phase.
  4. Wattless current cannot flow in purely inductive circuits.
Show answer and explanation

Answer: (A) The current is completely wattless as the phase difference between voltage and current is π/2.

In a purely inductive circuit, the current lags the voltage by a phase angle of π/2. The power factor cos ϕ = cos(π/2) = 0, so the average power consumed over a full cycle is zero. Hence, the current in the circuit is purely wattless.

hard

6. A step-down transformer has an efficiency of 85%. It reduces 2200 V to 220 V to operate a device having an impedance of 22 Ω. Assuming a purely resistive load, the current drawn by the primary coil from the mains is:

  1. 10.0 A
  2. 0.85 A
  3. 1.00 A
  4. 1.18 A
Show answer and explanation

Answer: (D) 1.18 A

Secondary voltage V_s = 220 V and R = 22 Ω, so secondary current I_s = 220 / 22 = 10 A. Output power P_out = V_s × I_s = 220 × 10 = 2200 W. With efficiency η = 0.85, input power P_in = P_out / η = 2200 / 0.85 ≈ 2588.2 W. Since P_in = V_p × I_p, the primary current is I_p = 2588.2 / 2200 = 1 / 0.85 ≈ 1.18 A.

easy

7. An ideal transformer has 500 primary turns and 4000 secondary turns. If 440 V AC is applied to the primary, the secondary voltage is:

  1. 1.76 × 10⁶ V
  2. 3520 V
  3. 440 V
  4. 55 V
Show answer and explanation

Answer: (B) 3520 V

Vs/Vp = Ns/Np, so Vs = 440 × 4000/500 = 3520 V.

easy

8. An ideal transformer has 200 primary turns and 2000 secondary turns. If 440 V AC is applied to the primary, the secondary voltage is:

  1. 440 V
  2. 44 V
  3. 8.8 × 10⁵ V
  4. 4400 V
Show answer and explanation

Answer: (D) 4400 V

Vs/Vp = Ns/Np, so Vs = 440 × 2000/200 = 4400 V.

easy

9. In an alternating current circuit, an alternating voltage V = 200 sin(ωt + π/3) V is applied, resulting in a current I = 10 sin(ωt - π/6) A. The wattless component of the current is:

  1. 5 A
  2. 5√2 A
  3. 5√3 A
  4. 10 A
Show answer and explanation

Answer: (B) 5√2 A

The phase difference between voltage and current is φ = (π/3) - (-π/6) = π/2 rad. The peak value of current is I₀ = 10 A, so I_rms = 10/√2 = 5√2 A. Since φ = 90°, sin φ = 1. Therefore, the wattless component is I_rms sin φ = 5√2 × 1 = 5√2 A.

easy

10. An ideal transformer has 1000 primary turns and 100 secondary turns. If 440 V AC is applied to the primary, the secondary voltage is:

  1. 440 V
  2. 44000 V
  3. 44 V
  4. 4400 V
Show answer and explanation

Answer: (C) 44 V

Vs/Vp = Ns/Np, so Vs = 440 × 100/1000 = 44 V.

easy

11. An ideal transformer has 200 primary turns and 250 secondary turns. If 220 V AC is applied to the primary, the secondary voltage is:

  1. 55000 V
  2. 220 V
  3. 275 V
  4. 176 V
Show answer and explanation

Answer: (C) 275 V

Vs/Vp = Ns/Np, so Vs = 220 × 250/200 = 275 V.

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12. An electric power station produces 200 kW of power at 10 kV. The power is transmitted through transmission lines having a total resistance of 5 Ω to a substation. What is the power loss during transmission?

  1. 2000 W
  2. 400 W
  3. 1000 W
  4. 100 W
Show answer and explanation

Answer: (A) 2000 W

Current transmitted is I = P / V = 200 × 10³ W / (10 × 10³ V) = 20 A. The power lost as heat in the transmission wires is P_loss = I² R = (20)² × 5 = 400 × 5 = 2000 W (or 2 kW).

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