Transformers: Practice Problem & Solution
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: (2010 Pre)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Transformers. Understanding the underlying formula is key to arriving at the correct answer below:
Efficiency of a transformer is given by $\eta = \frac{V_s I_s}{V_p I_p}$.
Substituting the given values: $0.80 = \frac{440 \times 2.0}{220 \times I_p}$.
Rearranging for primary current gives $I_p = \frac{880}{220 \times 0.8} = \frac{4}{0.8} = 5.0 A$.
Detailed Solution (Best of Understanding Complex Concepts)
Calculating Primary Current of a Transformer with 80% Efficiency
Concept Involved: Transformer Efficiency and Power Transfer
The fundamental concept tested in this question is the efficiency of a real transformer and the relationship between voltage, current, and electrical power.
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Ideal vs. Real Transformers: In an ideal transformer, there are zero power losses, meaning the input power equals the output power. However, practical transformers experience energy losses (such as copper loss from winding resistance, eddy currents, and hysteresis).
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Electrical Power: The electrical power ($P$) in a circuit is the product of its voltage ($V$) and current ($I$).
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Input Power ($P_{in}$) = $V_p \times I_p$
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Output Power ($P_{out}$) = $V_s \times I_s$
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Transformer Efficiency ($\eta$): Efficiency is defined as the ratio of the useful output power to the total input power.$$\eta = \frac{P_{out}}{P_{in}} = \frac{V_s I_s}{V_p I_p}$$
Given Data:
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Input Voltage (Primary Voltage), $V_p = 220 \text{ V}$
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Output Voltage (Secondary Voltage), $V_s = 440 \text{ V}$
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Output Current (Secondary Current), $I_s = 2.0 \text{ A}$
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Efficiency of the transformer, $\eta = 80\% = 0.80$
Step-by-Step Calculation:
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Substitute the given values into the transformer efficiency formula:$$0.80 = \frac{440 \times 2.0}{220 \times I_p}$$
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Simplify the numerator to find the output power:$$0.80 = \frac{880}{220 \times I_p}$$
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Simplify the fraction:$$0.80 = \frac{4}{I_p}$$
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Rearrange the equation to isolate the primary current ($I_p$):$$I_p = \frac{4}{0.80}$$
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Calculate the final value:$$I_p = 5.0 \text{ A}$$
Conclusion:
To deliver $2.0 \text{ A}$ at $440 \text{ V}$ with an $80\%$ efficiency, the primary windings of the transformer must draw a current of $5.0 \text{ A}$. Therefore, the correct answer is Option 1.
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