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Chapter 21: Problem 43

If the rms current in a \(50 \mathrm{~Hz}\) AC circuit is \(5 \mathrm{~A}\), the value of the current \(1 / 300\) seconds after its value \mathrm{\\{} b e c o m e s ~ z e r o ~ i s ~ [NCERT Exemplar] (a) \(5 \sqrt{2} \mathrm{~A}\) (b) \(5 \sqrt{3 / 2} \mathrm{~A}\) (c) \(5 / 6 \mathrm{~A}\) (d) \(5 / \sqrt{2} \mathrm{~A}\)

Short Answer

Expert verified
(b) \(5 \sqrt{3/2} \mathrm{~A}\).

Step by step solution

01

Determine angular frequency

In an AC circuit, the angular frequency \( \omega \) is calculated from the frequency using \( \omega = 2 \pi f \). For a frequency \( f = 50 \mathrm{~Hz} \), the angular frequency is \( \omega = 2 \pi \times 50 = 100 \pi \mathrm{~rad/s} \).
02

Calculate the time elapsed after zero current

We are given that the time after the current becomes zero is \( \frac{1}{300} \mathrm{~s} \).
03

Expression for instantaneous current

The instantaneous current \( i(t) \) in an AC circuit can be expressed as \( i(t) = I_0 \sin(\omega t) \), where \( I_0 \) is the peak current. Since the current is zero at \( t = 0 \), the expression simplifies to \( i(t) = I_0 \sin(\omega t) \).
04

Relate rms current to peak current

The relationship between the RMS current \( I_{rms} \) and peak current \( I_0 \) is \( I_{rms} = \frac{I_0}{\sqrt{2}} \). Given \( I_{rms} = 5 \mathrm{~A} \), we can find \( I_0 = 5\sqrt{2} \mathrm{~A} \).
05

Evaluate the instantaneous current

Using \( i(t) = I_0 \sin(\omega t) \), substitute \( \omega = 100\pi \), \( t = \frac{1}{300} \) seconds, and \( I_0 = 5\sqrt{2} \). Evaluate \( i(t) = 5\sqrt{2} \sin(100\pi \times \frac{1}{300}) \).
06

Solve the sine function

Calculate \( \sin(\frac{100\pi}{300}) = \sin(\frac{\pi}{3}) \). Knowing \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \), substitute this value back into the expression for \( i(t) \).
07

Final calculation for instantaneous current

Evaluate \( i(t) = 5\sqrt{2} \times \frac{\sqrt{3}}{2} = 5 \times \sqrt{\frac{3}{2}} \). The answer is \( 5\sqrt{\frac{3}{2}} \mathrm{~A} \).

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Key Concepts

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

RMS Current
In AC circuits, RMS current is a crucial concept. It stands for "Root Mean Square" current and represents a sort of average current value over time. While the current in AC systems constantly alternates, RMS gives us a consistent value. If a DC circuit was to produce the same amount of heat as the AC circuit in a resistor, it would have to carry this RMS current value.
The formula to find the RMS current (\(I_{rms}\)) from the peak current (\(I_0\)) is:
  • \(I_{rms} = \frac{I_0}{\sqrt{2}}\)
When the RMS current in the given problem is 5 A, it implies the AC circuit has an intensity equivalent to this, over time in a resistor-load standpoint.
Angular Frequency
Angular frequency (\(\omega\)) is a vital parameter that indicates how fast an alternating current oscillates. In our electrical system, it is related to regular frequency (\(f\)) by the equation:
  • \(\omega = 2\pi f\)
Measured in radians per second, angular frequency helps in understanding circuit behaviors. It determines the speed of oscillation. In this problem, if the frequency is 50 Hz, the angular frequency is calculated as:
  • \(\omega = 2 \pi \times 50 = 100 \pi \text{ rad/s}\)
This tells us that the current completes 100 full cycles in a 2\(\pi\) second interval, offering insight into the periodic motion pattern.
Peak Current
The peak current (\(I_0\)) refers to the maximum amplitude that the current reaches during its oscillations in an AC circuit. It's the highest possible current value and directly corresponds to the peak voltage in the system. To find the peak current from RMS current, we use:
  • \(I_{0} = I_{rms} \times \sqrt{2}\)
In the context of this exercise, if the RMS current is 5 A, the peak current computes as:
  • \(I_0 = 5 \times \sqrt{2} = 5\sqrt{2} \text{ A}\)
This gives us the maximum current occurring along each cycle.
Instantaneous Current
Instantaneous current is a snapshot of the current's value at any particular time in an AC circuit. It varies with time and can be calculated using the formula:
  • \(i(t) = I_0 \sin(\omega t)\)
This expression shows how the current flows and changes at any time during the cycle. Let's break down its use in our problem. Given that the peak current and angular frequency are known, if at time \(t = \frac{1}{300}\) s we need to find \(i(t)\), it becomes:
  • \(i(t) = 5\sqrt{2} \sin\left(100\pi \times \frac{1}{300}\right)\)
Further simplification uses:
  • \(\sin\left(\frac{100\pi}{300}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\)
Thus, the instantaneous current evaluates to \(5\sqrt{\frac{3}{2}} \text{ A}\), showing the current's magnitude at that exact moment.

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Most popular questions from this chapter

What is the self-inductance of an air core solenoid \(1 \mathrm{~m}\) long, diameter \(0.05 \mathrm{~m}\), if it has 500 turns? Take \(\pi^{2}=10\) (a) \(3.15 \times 10^{-4} \mathrm{H}\) (b) \(4.8 \times 10^{-4} \mathrm{H}\) (c) \(5 \times 10^{-4} \mathrm{H}\) (d) \(6.25 \times 10^{-4} \mathrm{H}\)

A coil is wound on a core of rectangular cross-section. If all the linear dimensions of core are increased by a factor 2 and number of turns per unit length of coil remains same, the self-inductance increases by a factor of (a) 16 (b) 8 (c) 4 (d) 2

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If the self-inductance of 500 turn coil is \(125 \mathrm{mH}\), then the self- inductance of similar coil of 800 turns is [Kerala CET 2008] (a) \(48.8 \mathrm{mH}\) (b) \(200 \mathrm{mH}\) (c) \(187.5 \mathrm{mH}\) (d) \(320 \mathrm{mH}\)

What is the self-inductance of an air core solenoid \(1 \mathrm{~m}\) long, diameter \(0.05 \mathrm{~m}\), if it has 500 turns? Take \(\pi^{2}=10\) (a) \(3.15 \times 10^{-4} \mathrm{H}\) (b) \(4.8 \times 10^{-4} \mathrm{H}\) (c) \(5 \times 10^{-4} \mathrm{H}\) (d) \(6.25 \times 10^{-4} \mathrm{H}\)
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