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Difficulty: HardElectromagnetic Induction

A step-down transformer connected to a 240 V240\text{ V} AC mains supply operates a 12 V,48 W12\text{ V}, 48\text{ W} lamp at its normal brightness rating. If the efficiency of the transformer is 80%80\%, what is the electric current drawn by the primary winding from the mains supply?

  1. A
    0.16 A0.16\text{ A}
  2. B
    0.20 A0.20\text{ A}
  3. 0.25 A0.25\text{ A}Answer
  4. D
    4.00 A4.00\text{ A}

Answer

0.25 A0.25\text{ A}
The output power delivered to the lamp is 48 W48\text{ W}. Accounting for the transformer's 80%80\% efficiency, the input power required at the primary winding is 48 W0.80=60 W\frac{48\text{ W}}{0.80} = 60\text{ W}. Since the primary voltage is 240 V240\text{ V}, the primary current drawn is Ip=60 W240 V=0.25 AI_p = \frac{60\text{ W}}{240\text{ V}} = 0.25\text{ A}.

Step-by-Step Solution

1
Determine the power output at the secondary winding
Ps=48 WP_s = 48\text{ W}
The lamp operates at its normal rating, so secondary power equals the lamp rating.
2
Calculate the required power input to the primary winding using transformer efficiency
Pp=Psη=48 W0.80=60 WP_p = \frac{P_s}{\eta} = \frac{48\text{ W}}{0.80} = 60\text{ W}
Efficiency is defined as η=PsPp\eta = \frac{P_s}{P_p}, meaning primary input power must exceed secondary output power due to losses.
3
Compute the primary current drawn from the supply
Ip=PpVp=60 W240 V=0.25 AI_p = \frac{P_p}{V_p} = \frac{60\text{ W}}{240\text{ V}} = 0.25\text{ A}
Power in an AC primary circuit is given by Pp=VpIpP_p = V_p I_p assuming a purely resistive secondary load.

Key Concept

Transformer Efficiency and Power Transfer in Electromagnetic Induction
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Topics

Electromagnetic Induction

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