Problem 62 - Master Resource Book in JEE Main Physics

Question

A transformer is used to light \(140 \mathrm{~W}, 24 \mathrm{~V}\) lamp from \(240 \mathrm{~V}\) AC mains, The current in the mains is \(0.7 \mathrm{~A}\). The efficiency of transformer is nearest to (a) 909 (b) 8096 (c) 7096 (d) \(60 \%\)

Step-by-Step Solution

Answer

The transformer's efficiency is closest to (b) 8096.

Step 1: Determine Input Power

The input power to the transformer on the primary side can be calculated using the formula: \( P_{in} = V_{in} \times I_{in} \). Here, \( V_{in} = 240 \text{ V} \) and \( I_{in} = 0.7 \text{ A} \). Thus, \( P_{in} = 240 \times 0.7 = 168 \text{ W} \).

Step 2: Determine Output Power

The output power on the secondary side of the transformer is the power of the lamp, which is given as \( P_{out} = 140 \text{ W} \).

Step 3: Calculate Efficiency

The efficiency of the transformer \( \eta \) is calculated using the formula: \( \eta = \left( \frac{P_{out}}{P_{in}} \right) \times 100 \% \). Substituting the values, we get \( \eta = \left( \frac{140}{168} \right) \times 100 \% \approx 83.33\% \).

Step 4: Match with Given Options

Compare the calculated efficiency of approximately \( 83.33\% \) with the given options. Although \( 83.33\% \) is not exactly given, the closest listed percentage option is \( 80\% \).

Key Concepts

Transformer Equations

Transformers play a crucial role in electrical circuits by stepping voltage levels up or down based on requirements. Understanding the fundamental equations of transformers is key to grasping their functionality. The primary equation associated with transformers is based on the power input to the primary coil and the power output from the secondary coil. This is represented as:

• Primary power: \(P_{in} = V_{in} \times I_{in}\)
• Secondary power: \(P_{out} = V_{out} \times I_{out}\)

In an ideal transformer, where there are no losses, the input power should equal the output power, meaning \(P_{in} = P_{out}\). However, in real-world applications, efficiency must be considered due to energy losses caused by various factors like heat and resistance.
The efficiency, \(\eta\), can be calculated using the equation:\[\eta = \left(\frac{P_{out}}{P_{in}}\right) \times 100\%\]This formula helps in determining how effectively a transformer transfers electrical energy without losses.

Power Calculations

Power calculations in transformers are essential for determining how much energy is moved from one circuit to another. In an AC circuit, the power is calculated using voltage and current values. The conventional formula for power \(P\) is given by:\[P = V \times I\]where:

• \(V\) is the voltage across the circuit
• \(I\) is the current flowing through the circuit

For example, if a transformer input has a voltage of \(240 \text{ V}\) and a current of \(0.7 \text{ A}\), the input power \(P_{in}\) is:\[P_{in} = 240 \times 0.7 = 168 \text{ W}\]Power calculations not only help determine the workload of electrical devices but also aid in identifying efficiency levels within a transformer, ensuring it functions correctly and maintains energy consumption standards.

AC Circuits

AC (Alternating Current) circuits are a type of electrical circuit in which the current changes direction periodically. In most homes and industries, AC is the dominant form of electricity for reasons including easier transmission over long distances and the ability to transform voltage levels using transformers effectively.
An understanding of AC circuits is essential when dealing with transformers. Here, the sinusoidal wave form of AC current means that both voltage and current constantly vary over time. AC voltage levels are typically represented in terms of their effective value known as root mean square (RMS). For a sinusoidal waveform, RMS voltage \(V_{rms}\) is calculated from the peak voltage \(V_{peak}\):\[V_{rms} = \frac{V_{peak}}{\sqrt{2}}\]This RMS value is what is usually used in practical power calculations in AC circuits because it equivalently represents the DC level of power supply that would deliver the same average power.