Problem 68 - Master Resource Book in JEE Main Physics

Question

A transformer is used to light a \(100 \mathrm{~W}-110 \mathrm{~V}\) lamp from \(220 \mathrm{~V}\) mains. If main current is \(0.5 \mathrm{~A}\), efficiency of transformer is (a) \(90 \%\) (b) 9596 (c) 9696 (d) \(99 \%\)

Step-by-Step Solution

Answer

The efficiency of the transformer is approximately 90% (option a).

Step 1: Understanding Transformer Power Equation

The power input to the transformer can be described as the product of voltage and current from the power mains. For this transformer, with a voltage of 220 V and a current of 0.5 A, the power input can be computed as follows: \[ P_{in} = V_{in} \times I_{in} = 220 \times 0.5 = 110 \, \text{W} \]

Step 2: Finding Power Output

The power output required by the transformer is the power consumed by the lamp, which is given directly as 100 W.

Step 3: Calculating Efficiency

The efficiency of a transformer is defined by the ratio of output power to input power, multiplied by 100 to get a percentage. Therefore, we calculate efficiency using:\[ \text{Efficiency} = \left( \frac{P_{out}}{P_{in}} \right) \times 100 \%\]Substitute the known values:\[ \text{Efficiency} = \left( \frac{100}{110} \right) \times 100 \% \approx 90.91 \% \]

Step 4: Selecting Closest Answer

Given the options, the closest value to the calculated efficiency of approximately 90.91% is option (a) 90%.

Key Concepts

Power Input

When talking about transformers, power input is a fundamental concept. It basically represents the total power absorbed by the transformer from the power mains or source. To determine this, we simply need to measure two things: the voltage across the primary coil (\( V_{in} \)) and the current flowing through it (\( I_{in} \)). The relationship is represented by the formula:\[ P_{in} = V_{in} \times I_{in} \]In our exercise, the transformer receives a voltage of 220 V and a current of 0.5 A. Applying these values into the formula gives us:\[ P_{in} = 220 \times 0.5 = 110 \, \text{W} \]This means the transformer takes in 110 watts from the power source. It's crucial to grasp power input clearly as it establishes the basis for calculating efficiency later.

Power Output

Power output is another crucial aspect of transformer functionality. It indicates the amount of power delivered to the load—in this case, a lamp. For transformers, the power output is usually measured at the secondary coil and reflects how much energy the device consuming the power effectively receives. - **Given in the Exercise**: The lamp is specified to require a power of 100 W. - **Contextual Understanding**: Power output comprises the useful energy flowing out of the transformer, specifically converted for usage by the lamp. Output power indicates real-world performance. By comparing it to the input power, we can derive the efficiency of the transformer system.

Efficiency Calculation

Efficiency in transformers relates to how effectively the device converts input power to output power. It indicates the percentage of input power that is successfully transformed to useful output. Calculating efficiency is straightforward:- **Formula**: The efficiency (\( \eta \)) is calculated by: \[ \text{Efficiency} = \left( \frac{P_{out}}{P_{in}} \right) \times 100\% \]Substituting the known values into this formula for our particular problem:\[ \text{Efficiency} = \left( \frac{100}{110} \right) \times 100\% \approx 90.91\% \]This value, approximately 90.91%, illustrates how efficiently the transformer converts its input to output power—a pivotal metric for assessing performance.- **Selection of Closest Answer**: The closest answer to 90.91% efficiency provided in the multiple choices is 90%, thereby making it the correct choice. Understanding this calculation is vital for anyone looking to evaluate or select transformers based on their performance attributes.