Question

The RMS value of ac of $50Hz$ is $10A$. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be:

• A. $2x{10}^{-2}s$ and $14.14A$
• B. $1x{10}^{-2}s$ and $7.07A$
• C. $5x{10}^{-3}s$ and $7.07A$
• D. $5x{10}^{-3}s$ and $14.14A$

Answer 1

The correct option is D

$5x{10}^{-3}s$ and $14.14A$

Step 1: Given Data

${\mathrm{I}}_{\mathrm{rms}}=10\mathrm{A};{\mathrm{f}}_{\mathrm{rms}}=50\mathrm{Hz}$

Step 2: Formula used

$\mathrm{T}=\frac{1}{{\mathrm{f}}_{\mathrm{rms}}}$

${\mathrm{I}}_{\mathrm{o}}=\sqrt{2}{\mathrm{I}}_{\mathrm{rms}}$

Let $I_0$ be the peak value of the current.

$T$ is the time period of the AC function.

$f_{rms}$ is the r.m.s value of frequency

$I_{rms}$ is the r.m.s value of current

Time taken by the alternating current to reach maximum from zero is $\frac{T}{4}$

Step 3: Calculate the time to reach the maximum peak value

The time period is the time taken for one whole wave to pass.

One whole wave consists of a crest and a trough.

One-fourth of this wave can be considered the distance between zero and the first peak. Hence it takes one-fourth of the time to reach the peak

Hence, the time taken to reach the maximum peak from zero is =$\frac{T}{4}$

$\mathrm{T}=\frac{1}{{\mathrm{f}}_{\mathrm{rms}}}$

Time taken to reach the maximum is, $\frac{1}{4{\mathrm{f}}_{\mathrm{rms}}}=\frac{1}{4\times 50}=\frac{1}{200}=5\times {10}^{-3}\mathrm{s}$

Step 4: Calculate the peak value of current

$$\begin{gathered} {\mathrm{I}}_{\mathrm{o}}=\sqrt{2}{\mathrm{I}}_{\mathrm{rms}} \\ {\mathrm{I}}_{\mathrm{o}}=\sqrt{2}\times 10 \\ {\mathrm{I}}_{\mathrm{o}}=1.414\times 10\mathrm{A} \\ {\mathrm{I}}_{\mathrm{o}}=14.14\mathrm{A} \end{gathered}$$

Thus, the time taken for reaching the peak value of current from zero is $5\times {10}^{-3}\mathrm{s}$ and the peak value of current is $14.14A$, and therefore the correct answer is option (D).