Examkin

Question

Difficulty: HardInductors and Transformers

An iron-core step-down transformer connected to a 240 V AC240\text{ V AC} power line has 1200 turns1{}200\text{ turns} in its primary winding and 150 turns150\text{ turns} in its secondary winding. The secondary winding is connected across a 15 Ω15\ \Omega resistive load. Assuming an ideal transformer with no power losses, what is the current drawn by the primary winding from the AC source?

  1. A
    16.0 A16.0\text{ A}
  2. 0.25 A0.25\text{ A}Answer
  3. C
    2.00 A2.00\text{ A}
  4. D
    0.025 A0.025\text{ A}

Answer

0.25 A0.25\text{ A}
The transformer has an 8:18:1 step-down turns ratio (1200/150=81{}200 / 150 = 8). The secondary voltage is 240 V/8=30 V240\text{ V} / 8 = 30\text{ V}. Connecting a 15 Ω15\ \Omega resistor across 30 V30\text{ V} produces a secondary current of 30 V/15 Ω=2.0 A30\text{ V} / 15\ \Omega = 2.0\text{ A}. Since power in an ideal transformer is conserved (Pp=PsP_p = P_s), the primary current is stepped down by the same ratio: 2.0 A/8=0.25 A2.0\text{ A} / 8 = 0.25\text{ A}.

Step-by-Step Solution

1
Calculate the turns ratio (aa) of the transformer.
a=NpNs=1200 turns150 turns=8a = \frac{N_p}{N_s} = \frac{1{}200\text{ turns}}{150\text{ turns}} = 8
The ratio of primary to secondary turns governs the voltage step-down and current step-up proportions.
2
Calculate the secondary voltage (VsV_s) across the load.
Vs=Vpa=240 V8=30 VV_s = \frac{V_p}{a} = \frac{240\text{ V}}{8} = 30\text{ V}
In a step-down transformer, secondary voltage equals primary voltage divided by the turns ratio.
3
Calculate the secondary current (IsI_s) through the load resistor using Ohm's Law.
Is=VsRL=30 V15 Ω=2.0 AI_s = \frac{V_s}{R_L} = \frac{30\text{ V}}{15\ \Omega} = 2.0\text{ A}
Current through a resistive load is directly proportional to voltage and inversely proportional to resistance.
4
Calculate the primary current (IpI_p) drawn from the AC source.
Ip=Isa=2.0 A8=0.25 AI_p = \frac{I_s}{a} = \frac{2.0\text{ A}}{8} = 0.25\text{ A}
Conservation of power in an ideal transformer requires that VpIp=VsIsV_p I_p = V_s I_s, meaning primary current is the secondary current divided by the turns ratio.

Key Concept

Ideal transformer transformation ratios (Np/Ns=Vp/Vs=Is/IpN_p / N_s = V_p / V_s = I_s / I_p)

Alternative Method

Using reflected impedance: The load resistance reflected to the primary side is Rp′=a2×RL=82×15 Ω=64×15 Ω=960 ΩR_p' = a^2 \times R_L = 8^2 \times 15\ \Omega = 64 \times 15\ \Omega = 960\ \Omega. The primary current is then directly calculated using Ohm's Law: Ip=VpRp′=240 V960 Ω=0.25 AI_p = \frac{V_p}{R_p'} = \frac{240\text{ V}}{960\ \Omega} = 0.25\text{ A}.
Estimated Time:2m 0s
Rate this question

Topics

Inductors and Transformers

More questions from this topic