Question

An A.C source of voltage $V=220\sin (100\pi t)volts$ is connected with $R=50\Omega$. The time interval in which the current goes from its peak value to half of the peak value is

• A. $\begin{array}{l}\frac{1}{400}\sec \end{array}$
• B. $\begin{array}{l}\frac{1}{50}\sec \end{array}$
• C. $\begin{array}{l}\frac{1}{300}\sec \end{array}$
• D. $\begin{array}{l}\frac{1}{200}\sec \end{array}$

Answer 1

The correct option is C

$\begin{array}{l}\frac{1}{300}\sec \end{array}$

Step 1: Given data

Voltage, $V=220\sin \left(100\pi t\right)volts$

Resistance,$R=50\Omega$

Step 2: Finding the value of current, $I$:

$I=\frac{V}{R}$

$I=\frac{220}{50}\sin \left(100\pi t\right)$……….$\left(1\right)$

Step 3: Time is taken by the current to go from its peak value to half of its peak value:

Comparing the equation $\left(1\right)$ with the standard equation,

$I=I_{max}\sin \left(\omega t\right)$

Thus, $$\begin{gathered} \omega =2\pi f \\ =100\pi \end{gathered}$$ [$\omega$ is the angular frequency, $f$ is ordinary frequency, $t$ is time]

$f=50Hz$

For oscillation to go from its peak value to half of its peak value, the time interval is $∆T$ which is $\frac{T}{6}$

$T=\frac{1}{f}$

$∆T=\frac{T}{6}$

$$\begin{gathered} =\frac{1}{6xf} \\ =\frac{1}{6x50} \\ =\frac{1}{300}sec. \end{gathered}$$

Hence, the option $\left(C\right)$ is correct