Problem 67 - Master Resource Book in JEE Main Physics

Question

An AC voltage source has an output of \(\Delta V=(200 \mathrm{~V}) \sin 2 \pi f t\). This source is connected to a \(100 \mathrm{~W}\) resistor. Rms current in the resistance is [Kerala CET 2008] (a) \(1.41 \mathrm{~A}\) (b) \(2.41 \mathrm{~A}\) (c) \(3.41 \mathrm{~A}\) (d) \(0.71 \mathrm{~A}\)

Step-by-Step Solution

Answer

The rms current in the resistor is 0.71 A (option d).

Step 1: Identify Given Values

We are given the peak voltage \( V_0 = 200 \text{ V} \) and the resistor power \( P = 100 \text{ W} \). We need to find the rms current \( I_{rms} \).

Step 2: Calculate RMS Voltage

The rms voltage \( V_{rms} \) is calculated from the peak voltage using the formula: \( V_{rms} = \frac{V_0}{\sqrt{2}} \). Flipping the numbers in: \( V_{rms} = \frac{200}{\sqrt{2}} \approx 141.42 \text{ V} \).

Step 3: Use Power Formula for Resistor

The power dissipated by the resistor is given by the formula: \( P = V_{rms} \times I_{rms} \). We need to solve this formula for \( I_{rms} \).

Step 4: Solve for RMS Current

Rearrange the power formula to find \( I_{rms} \): \( I_{rms} = \frac{P}{V_{rms}} = \frac{100}{141.42} \approx 0.71 \text{ A} \).

Step 5: Determine Correct Option

The calculated \( I_{rms} \) of \( 0.71 \text{ A} \) matches option (d).

Key Concepts

AC Voltage Source

An AC voltage source is a fundamental component in electrical circuits that provides an alternating voltage. Unlike direct current (DC) sources that provide a constant output, AC sources supply voltage that varies with time, typically in a sinusoidal form.
The formula for an AC voltage source is often given as\[ \Delta V = V_0 \sin(2 \pi f t) \]where:

• \( \Delta V \) is the instantaneous voltage at time \( t \)
• \( V_0 \) is the peak voltage
• \( f \) is the frequency of the AC source
• \( t \) is time

This time-dependent nature allows AC voltage to be used for powering many types of equipment and making long-distance power transmission economically viable due to reduced energy loss.

Power in Resistor

The power dissipated by a resistor in an electrical circuit is crucial for understanding how much energy is being converted into heat within the resistor. The formula that relates the power \( P \) to the root mean square (rms) voltage and current is\[ P = V_{rms} \times I_{rms} \]This equation indicates the relationship between power, voltage, and current in AC circuits.
For resistors, the power can also be expressed as:

• \( P = I_{rms}^2 \times R \)
• \( P = \frac{V_{rms}^2}{R} \)

Here \( R \) is the resistance measured in ohms. Understanding this relationship helps in determining how much current can flow safely through the resistor without causing overheating.

RMS Voltage

RMS Voltage, or root mean square voltage, is a statistical measure of the magnitude of a varying voltage. It provides a consistent representation of equivalent DC voltage that would deliver the same power to a resistor.
The formula to calculate the rms voltage \( V_{rms} \) from the peak voltage \( V_0 \) is:\[ V_{rms} = \frac{V_0}{\sqrt{2}} \]This equation arises because the power delivered by the AC voltage source oscillates and we are interested in the effective value that translates into real work or heat. By using rms voltage, we can directly apply the basic power formula applicable to constant DC.The idea is straightforward: rms voltage simplifies the analysis of AC circuits by providing a single value that depicts the average power delivery capability of the alternating waveform.

Root Mean Square

The concept of root mean square (RMS) finds its prominence in a variety of fields, especially in AC electrical circuits. It is a statistical measure used to determine the magnitude of a set of values. For AC voltage and current, RMS values represent the effective level of those waveforms.
The RMS value of a sinusoidal waveform like voltage or current is given by:\[ X_{rms} = \sqrt{ \frac{1}{T} \int_0^T x^2(t) \, dt } \]This formula, when applied to a sine wave, simplifies to the well-known results for rms voltage and current when comparing their peak values to rms values.
Understanding RMS is critical because:

• It allows for the comparison between AC and DC effectively.
• It provides an accurate description of energy transfer capabilities.
• It's integral in designing systems that are efficient and safe.

RMS makes calculations easier and more practical for engineers and helps convey real-world implications of AC systems.