Question

An AC source rated $220\text{V}$, $50\text{Hz}$ is connected to a resistor. The time taken by the current to change from its maximum to the rms value is:

• A. $0.25\text{ms}$
• B. $25\text{ms}$
• C. $2.5\text{ms}$
• D. $2.5\text{s}$

Answer 1

The correct option is C $2.5\text{ms}$

Given, $V_{rms}=200\text{V}$ and $f=50\text{Hz}$
So, $V_0=220\sqrt{2}\text{V}$ and $\omega =2\pi f=100\pi$

$\Rightarrow V=V_0\sin \omega t=220\sqrt{2}\sin \left(100\pi t\right)$

Current in the resistor, $I=I_0\sin 100\pi t$ where $I_0=\frac{220\sqrt{2}}{R}$

Let current is maximum for first time at time $t_1$:

$I_0=I_0\sin \left(100\pi t_1\right)$

$1=\sin \left(100\pi t_1\right)$

$\frac{\pi }{2}=100\pi t_1$

$t_1=\frac{1}{200}$

Suppose that current is at its RMS value for first time at time $t_2$:

$\frac{I_0}{\sqrt{2}}=I_0\sin \left(100\pi t_2\right)$

$\frac{1}{\sqrt{2}}=\sin \left(100\pi t_2\right)$

$\frac{\pi }{4}=100\pi t_2$

$t_2=\frac{1}{400}$

$\Delta t=t_1-t_2=\frac{1}{200}-\frac{1}{400}$

$\Delta t=\frac{1}{400}\text{s}=2.5\text{ms}$