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Step 1: Identify the Physical Situation
We have a resistor (of resistance R ) and an inductor (of inductance L ) connected in series to a battery of emf E . We are asked to find the numerical value of the ratio \frac{b}{a} when the maximum rate of energy storage in the inductor is given by \frac{E^a}{2b}\,\text{J/s} .
Step 2: Recall Energy Stored in an Inductor
The energy stored in an inductor carrying current I is:
E_{\text{inductor}} = \frac{1}{2}\,L\,I^2.
Step 3: Rate of Change of Energy (Power) in the Inductor
The instantaneous power going into the inductor is the time derivative of the stored energy:
P_{\text{inductor}}(t)
= \frac{d}{dt}\Bigl(\tfrac{1}{2} \, L \, I^2\Bigr)
= L\,I\,\frac{dI}{dt}.
Step 4: Current in the R–L Circuit
When a resistor R and inductor L are connected in series to a battery of emf E , the current as a function of time is:
I(t) = \frac{E}{R}\Bigl(1 - e^{-\tfrac{R}{L}\,t}\Bigr).
Its time derivative is:
\frac{dI}{dt} = \frac{E}{L}\,e^{-\tfrac{R}{L}\,t}.
Step 5: Substitute into the Power Expression
Substitute I(t) and \tfrac{dI}{dt} into P_{\text{inductor}}(t) = L\,I\,\tfrac{dI}{dt} :
P_{\text{inductor}}(t)
= L \,\Bigl(\frac{E}{R}\bigl(1 - e^{-\tfrac{R}{L}\,t}\bigr)\Bigr)
\Bigl(\frac{E}{L}\,e^{-\tfrac{R}{L}\,t}\Bigr).
Simplifying,
P_{\text{inductor}}(t)
= \frac{E^2}{R}\,\Bigl(1 - e^{-\tfrac{R}{L}\,t}\Bigr)\,e^{-\tfrac{R}{L}\,t}.
Step 6: Find the Time for the Maximum Power
To find when P_{\text{inductor}}(t) is maximum, note that it involves the product
\bigl(1 - x\bigr)\,x, where x = e^{-\frac{R}{L} t}.
The product (1 - x)x attains its maximum at x = \tfrac{1}{2}. Hence,
e^{-\tfrac{R}{L}\,t} = \frac{1}{2}
\quad\Longrightarrow\quad
t = \frac{L}{R}\,\ln(2).
Step 7: Maximum Rate of Energy Storage
Substituting x = \tfrac{1}{2} back in,
P_{\text{inductor, max}}
= \frac{E^2}{R}\,\Bigl(\tfrac{1}{2}\Bigr)\,\Bigl(\tfrac{1}{2}\Bigr)
= \frac{E^2}{4\,R}.
Thus, the maximum power stored in the inductor is \tfrac{E^2}{4R}\,\text{J/s} .
Step 8: Compare with the Given Form \frac{E^a}{2b}
We are given that the maximum rate has the form \frac{E^a}{2b}\,\text{J/s} . Comparing:
\frac{E^2}{4\,R} = \frac{E^a}{2b}.
Matching powers of E shows a = 2. Then comparing coefficients:
\frac{1}{4R} = \frac{1}{2b}
\quad\Longrightarrow\quad
2b = 4R
\quad\Longrightarrow\quad
b = 2R.
Step 9: Evaluate \frac{b}{a}
Since a = 2 and b = 2R , we get:
\frac{b}{a} = \frac{2R}{2} = R.
If the given resistor is 25\,\Omega, then
\tfrac{b}{a} = 25.
