Answer
(a) greater than
(b) $0.82A $
(c) $0.54A $
Work Step by Step
(a) We know that
$ I_{\circ}=\frac{V}{R}$
$ I_{\circ}=\frac{9.0V}{11\Omega}$
$ I_{\circ}=0.82A $
When the switch is closed then
$ I=\frac{9.0V}{11\Omega+5.6\Omega}$
$ I=0.54A $
Thus, the value of $ I\lt I_{\circ}$. That is, the initial current is greater than the final current.
(b) We know that
$ I_{\circ}=\frac{V}{R}$
$ I_{\circ}=\frac{9.0V}{11\Omega}$
$ I_{\circ}=0.82A $
(c) The current in the battery long after the switch is closed is:
$ I=\frac{V}{11\Omega+5.6\Omega}$
$ I=0.54A $
