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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) \(60 \%\)

Short Answer

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

Step by step solution

01

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, \( P_{in} = 240 \times 0.7 = 168 \text{ W} \).
02

Determine Output Power

The output power on the secondary side of the transformer is the power of the lamp, which is given as \( P_{out} = 140 \text{ W} \).
03

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 \( \eta = \left( \frac{140}{168} \right) \times 100 \% \approx 83.33\% \).
04

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 \( 80\% \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A long solenoid with 15 turns per \(\mathrm{cm}\) has a small loop of area \(2.0 \mathrm{~cm}^{2}\) placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from \(2.0 \mathrm{~A}\) to \(4.0 \mathrm{~A}\) in \(0.1 \mathrm{~s}\), what is the induced emf in the loop while the current is changing? [NCERT] (a) \(7.5 \times 10^{6} \mathrm{~V}\) (b) \(8.5 \times 10^{6} \mathrm{~V}\) (c) \(7.5 \times 10^{4} \mathrm{~V}\) (d) \(7.5 \times 10^{5} \mathrm{~V}\)

A cylindrical bar magnet is rotated about its axis. A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then [NCERT Exemplar] (a) a direct current flows in the ammeter \(A\). (b) no current flows through the ammeter \(A\). (c) an altemating sinusoidal current flows through the ammeter \(A\) with a time period \(T=\frac{2 \pi}{\omega}\). (d) a time varying non-sinosoidal current flows through the ammeter \(A\).

There are two coils \(A\) and \(B\) as shown in figure. A current starts flowing in \(B\) as shown, when \(A\) is moved towards \(B\) and stops when A stops moving. The current in \(A\) is counterclockwise. \(B\) is kept stationary when \(A\) moves. We can infer that \(\quad\) [NCERT Exemplar] (a) there is a constant current in the clockwise direction in \(A\). (b) there is a varying curent in \(A\). (c) there is no current in \(A\). (d) there is a constant current in the counterclockwise direction in \(A\)

A solenoid has 2000 turns wound over a length of \(0.30 \mathrm{~m}\). The area of its cross-section is \(1.2 \times 10^{-3} \mathrm{~m}^{2}\). Around its central section, a coil of 300 turns is wound. If an initial current of \(2 \mathrm{~A}\) in the solenoid is reversed in \(0.25 \mathrm{~s}\), then the emf induced in the coil is equal to (a) \(6 \times 10^{-4} \mathrm{~V}\) (b) \(4.8 \times 10^{-2} \mathrm{~V}\) (c) \(6 \times 10^{-2} \mathrm{~V}\) (d) \(48 \mathrm{kV}\)

Two coils \(X\) and \(Y\) are placed in a circuit such that a current changes by \(2 \mathrm{~A}\) in coil \(X\) and magnetic flux change of \(0.4\) Wb occurs in \(Y\). The value of mutual inductance of the coils is (a) \(0.8 \mathrm{H}\) (b) \(0.2 \mathrm{~Wb}\) (c) \(0.2 \mathrm{H}\) (d) \(5 \mathrm{H}\)

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