Answer
(a) $0.075J $
(b) $0.14J $
(c) decrease
Work Step by Step
(a) We know that the resistance of the two resistors (which are parallel) is given as
$\frac{1}{R_{eq}}=\frac{1}{7.5\Omega}+\frac{1}{14\Omega}$
This simplifies to:
$ R_{eq}=4.88\Omega $
Now $ U=\frac{1}{2}L(\frac{\epsilon}{R_{eq}})^2(1-e^{-1})^2$
We plug in the known values to obtain:
$ U=\frac{1}{2}[\frac{(12V)^2(62\times 10^{-3}H)}{4.88\Omega}](1-e^{-1})^2$
$ U=0.075J $
(b) As $ U=\frac{1}{2}L(\frac{\epsilon}{R_{eq}})(1-e^{-2})^2$
We plug in the known values to obtain:
$ U=[\frac{(12V)^2(62\times 10^{-3}H)}{(4.88\Omega)^2}](1-e^{-2})^2$
$ U=0.14J $
(c) We know that the characteristic time is given as $ t=\frac{L}{R_{eq}}$. This equation shows that when the value of $ R $ increases, the equivalent resistance increases as well and as a result the value of the characteristic time $ t $ decreases.
