Answer
a) $ R\approx 40\Omega $
b) less than
Work Step by Step
(a) We know that $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R}=\frac{1}{7.5\Omega}+\frac{1}{R}$
Now $ U=\frac{1}{2}L\epsilon^2[\frac{1}{7.5\Omega}+\frac{1}{R}]^2$
This simplifies to:
$ R=[\sqrt{\frac{2u}{L\epsilon^2}}-\frac{1}{7.5\Omega}]^{-1}$
We plug in the known values to obtain:
$ R=[\sqrt{\frac{2u}{L\epsilon^2}}-\frac{1}{7.5\Omega}]^{-1}$
We plug in the known values to obtain:
$ R=[\sqrt{\frac{2(0.11J)}{62\times 10^{-3}H(12V)^2}}-\frac{1}{7.5\Omega}]^{-1}$
$ R\approx 40\Omega $
(b) We know that the energy stored in an inductor is inversely proportional to the square of the equivalent resistance. Thus, the value of $ R $ should be less to store more energy in the inductor.
