Question 200: AC fundamentals

Question

In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. $$\frac{E^{a}}{2 b} \mathrm{~J} / s$$ represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of $$\frac{b}{a}$$ will be __________.

Accepted Answer

$$
E=\frac{1}{2} L I^2
$$
Rate of energy storing $=\frac{d E}{d t}=L I \frac{d I}{d t}$
Now we Know for $R-L$ circuit
$$
I=\frac{E}{R}\left(1-e^{-t \frac{R}{L}}ight)
$$
So $\frac{d I}{d t}=\frac{E}{L} e^{-\frac{t R}{L}}$
$$
\frac{d E}{d t}=\frac{E^2}{R}\left(1-e^{-\frac{t R}{L}}ight)\left(e^{-t \frac{R}{L}}ight)
$$
Time at which rate of power storing will be $\max$
$$
\mathrm{t}=\frac{L}{R \ln 2}
$$
So $\frac{d E}{d t}=\frac{E^2}{R}\left(1-\frac{1}{2}ight) 	imes \frac{1}{2}$
$$
\Rightarrow \frac{E^2}{4 R}=\frac{E^2}{100}=\frac{E^2}{2 	imes 50}
$$
$$
a=2, b=50
$$
So $\frac{b}{a}=25$

AC fundamentals

In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. $\frac{E^{a}}{2 b} \mathrm{~J} / s represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of \frac{b}{a}$ will be __________.

Integer / Numeric Answer Type