Question

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 ______.
 

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.

Answer 1

Correct Answer: 25

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: \[ \frac{b}{a} = 25. \]