In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. \(\frac{E^a}{2b}\) 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 ______.
In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. \(\frac{E^a}{2b}\) 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 ______.
Show Hint
In RL circuits, the power dissipated in the resistor and stored in the inductor can be related using the energy storage formulas. Ensure the resistance and emf values are used consistently.
Correct Answer: 25
Solution and Explanation
Step 1: Power Supplied by the Battery
The power supplied by the battery is given by:
\[
P = \frac{E^2}{R},
\]
where \( E \) is the emf of the battery and \( R \) is the resistance. Substituting \( R = 25 \, \Omega \):
\[
P = \frac{E^2}{25}.
\]
Step 2: Power Stored in the Inductor
The maximum power stored in the inductor is given as:
\[
P_{\text{inductor}} = \frac{E^2}{2b}.
\]
Step 3: Relating \( b \) and \( a \)
Since the total resistance is \( R = 25 \, \Omega \), and the stored energy in the inductor is proportional to the power supplied:
\[
\frac{E^2}{25} = 2 \times \frac{E^2}{2b}.
\]
Simplify:
\[
b = 25.
\]
Step 4: Find the Ratio \( \frac{b}{a} \)
Since \( a = 1 \) (from standard proportionality), the ratio is:
\[
\frac{b}{a} = 25.
\]
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