Question

A $220\, volt$ input is supplied to a transformer. The output circuit draws a current of $2.0$ ampere at $440\, volts$. If the efficiency of the transformer is $80\%$, the current drawn by the primary windings of the transformer is

• A. 3.6 ampere
• B. 2.8 ampere
• C. 2.5 ampere
• D. 5.0 ampere

Answer 1

The Correct Option is D

To find the current drawn by the primary windings of the transformer, we need to use the efficiency of the transformer and the given parameters. Here's how we solve the problem step-by-step:

1. First, we know the formula for the efficiency of a transformer is given by: \[\text{Efficiency} = \frac{\text{Output Power}}{\text{Input Power}} \times 100\%\]
2. The output power of the transformer can be calculated using the formula: \[\text{Output Power} = V_{\text{secondary}} \times I_{\text{secondary}}\]
   • Given that V_{\text{secondary}} = 440\, \text{volts} and I_{\text{secondary}} = 2.0\, \text{A}, we have:
   • \text{Output Power} = 440\, \text{V} \times 2.0\, \text{A} = 880\, \text{W}
3. Using the efficiency value, we calculate the input power: \[\text{Input Power} = \frac{\text{Output Power}}{\text{Efficiency}}\]
   • The efficiency is 80\%\ or 0.8 in decimal.
   • \text{Input Power} = \frac{880\, \text{W}}{0.8} = 1100\, \text{W}
4. Now, we use the input power to find the primary current: \[\text{Input Power} = V_{\text{primary}} \times I_{\text{primary}}\]
   • Given V_{\text{primary}} = 220\, \text{volts},
   • I_{\text{primary}} = \frac{\text{Input Power}}{V_{\text{primary}}} = \frac{1100\, \text{W}}{220\, \text{V}} = 5.0\, \text{A}

Therefore, the current drawn by the primary windings of the transformer is 5.0 amperes. This is consistent with the provided correct answer.