Transformers: Practice Problem & Solution
A step down transformer connected to an ac mains supply of $220 V$ is made to operate at $11 V, 44 W$ lamp. Ignoring power losses in the transformer, what is the current in the primary circuit? (2021)
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:
For an ideal transformer, input power equals output power: $P_p = P_s$.
The output power (power of the lamp) is $P_s = 44 W$.
Primary voltage is $V_p = 220 V$, so $V_p I_p = 44$.
$I_p = \frac{44}{220} = 0.2 A$.
Detailed Solution (Best of Understanding Complex Concepts)
Concepts Involved:
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Ideal Transformer Principle: A transformer works on the principle of mutual induction. When power losses (such as flux leakage, eddy currents, copper loss, and hysteresis) are ignored, it is considered an "ideal transformer." In an ideal transformer, the input power at the primary coil is completely transferred to the output power at the secondary coil ($P_{\text{input}} = P_{\text{output}}$).
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Electric Power Formula: The electrical power ($P$) in a circuit is the product of its voltage ($V$) and current ($I$). Therefore, $P = VI$.
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Primary Power: $P_p = V_p \times I_p$
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Secondary Power: $P_s = V_s \times I_s$
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Step-by-Step Solution:
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Identify the given parameters:
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Voltage of the ac mains (Primary Voltage), $V_p = 220\text{ V}$
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Operating voltage of the lamp (Secondary Voltage), $V_s = 11\text{ V}$
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Power consumed by the lamp (Secondary Power), $P_s = 44\text{ W}$
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Apply the ideal transformer condition:
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Since the problem states to "ignore power losses," the power in the primary circuit must equal the power in the secondary circuit.
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$P_p = P_s$
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Therefore, the power in the primary circuit is $44\text{ W}$.
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Express primary power in terms of voltage and current:
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Using the power formula: $P_p = V_p \times I_p$
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Substitute the known values into the equation:
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$44 = 220 \times I_p$
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Solve for the primary current ($I_p$):
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$I_p = \frac{44}{220}$
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$I_p = \frac{1}{5}$
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$I_p = 0.2\text{ A}$
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The current in the primary circuit is $0.2\text{ A}$.
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