Physics 2 Marks Questions

In case of inductor:

$\text{V}-\text{L}\frac{\text{dI}}{\text{dt}}=0$

$\text{V}=\text{L}\frac{\text{dI}}{\text{dt}}$

$\int\text{dI}=\frac{\text{V}}{\text{L}}\int\text{dt}$

$\text{I}=\frac{\text{Vt}}{\text{L}}$

If direct current is connected across inductor current increases with time and transformer is also inductor. So, current can increase to large value and transfer can burn.

$\text{P}_\text{rms}=\frac{\text{P}_0}{2}=40\text{W.}$

Energy consumed $=\text{P}\times\text{t}=40\times100$

$=4000\text{J}=4.0\text{KJ}$

$\text{i}^2=\frac{1}{\tau}\int\limits_0^\tau\text{i}_0{^2}\text{e}^{\frac{-2\text{t}}{\tau}}\text{dt}$

$=\frac{\text{i}_0{^2}}{\tau}\int\limits_0^\tau\text{e}^{\frac{-2\text{t}}{\tau}}\text{dt}$

$=\frac{\text{i}_0{^2}}{\tau}\times\Big[\frac{\tau}{2}\text{e}^{\frac{-2\text{t}}{\tau}}\Big]_0^\tau$

$=-\frac{\text{i}_0{^2}}{\tau}\times\frac{\tau}{2}\big[\text{e}^{-2}-1\big]$

$\sqrt{\text{i}^2}=\sqrt{-\frac{\text{i}_0^2}{2}\Big(\frac{1}{\text{e}^2}-1\Big)}$

$=\frac{\text{i}_0^2}{2}\sqrt{\Big(\frac{\text{e}^2-1}{2}\Big)}$