Question

The peak value of an alternating current is 14.14 Amp., and its frequency is 50 Hz. Draw current-time graph for two cycles. Find r.m.s. value of current. What time will the current take to reach the peak value starting from zero ?

Hint:

In AC problems, values like 14.14, 1.414, or 0.707 are often hints to use \(\sqrt{2}\). Specifically, \(14.14 \approx 10\sqrt{2}\) and \(0.707 \approx 1/\sqrt{2}\). Recognizing these quickly can save a lot of calculation time.

Answer 1

Step 1: Understanding the Concept and Given Data:
This problem deals with a sinusoidal alternating current (AC). We are given the peak value (\(I_0\)) and the frequency (\(f\)) and need to find its RMS value, the time to reach the first peak, and visualize its waveform. \begin{itemize} \item Peak current, \(I_0 = 14.14\) A. \item Frequency, \(f = 50\) Hz. \end{itemize}

Step 2: Key Formulas and Calculations:
1. R.M.S. Value of Current (\(I_{rms}\)):
The relationship between the peak value and the RMS value of a sinusoidal current is: \[ I_{rms} = \frac{I_0}{\sqrt{2}} \] We are given \(I_0 = 14.14\) A. Note that \(14.14 \approx 10 \times 1.414 = 10\sqrt{2}\). \[ I_{rms} = \frac{10\sqrt{2}}{\sqrt{2}} = 10 \, \text{A} \] 2. Time to Reach the Peak Value:
The time period of one cycle is \(T = \frac{1}{f}\). \[ T = \frac{1}{50 \, \text{Hz}} = 0.02 \, \text{s} \] A sinusoidal current starts from zero, reaches its first positive peak at one-quarter of the time period (\(t = T/4\)). \[ t_{peak} = \frac{T}{4} = \frac{0.02 \, \text{s}}{4} = 0.005 \, \text{s} \]

Step 3: Current-Time Graph for Two Cycles:
We need to draw the graph for a time duration of \(2T = 2 \times 0.02 = 0.04\) s. The graph is a sine wave with an amplitude of 14.14 A. \begin{center} \begin{tikzpicture} \begin{axis}[ xlabel={Time (s)}, ylabel={Current (A)}, xmin=0, xmax=0.045, ymin=-16, ymax=16, axis lines=middle, grid=both, xtick={0, 0.005, 0.01, 0.015, 0.02, 0.025, 0.03, 0.035, 0.04}, xticklabels={0, T/4, T/2, 3T/4, T, 5T/4, 3T/2, 7T/4, 2T}, ytick={-14.14, 0, 14.14}, yticklabels={-14.14, 0, 14.14}, width=0.9\textwidth, height=0.5\textwidth, ] \addplot[smooth, thick, blue, domain=0:0.04, samples=200] {14.14*sin(2*pi*50*x*180/pi)}; \draw[dashed, gray] (axis cs:0.005,0) -- (axis cs:0.005,14.14); \draw[dashed, gray] (axis cs:0,14.14) -- (axis cs:0.005,14.14); \end{axis} \end{tikzpicture} \end{center}

Step 4: Final Answer: \begin{itemize} \item The r.m.s. value of the current is 10 A. \item The time taken for the current to reach the peak value starting from zero is 0.005 s. \end{itemize}