Answer
(a) $129V$
(b) decrease
Work Step by Step
(a) We know that
$V=(1.52A)(8.45\Omega+4.11\Omega)$
$V=19.091V$
Now $I_{13.8}=\frac{19.091V}{13.8\Omega}=1.38A$
$I_{17.2}=\frac{19.091V}{17.2\Omega}=1.11A$
Thus, the total current is given as
$I=1.38A+1.11A+1.52A=4.0A$
$V_{15\Omega}=(4.0A)(12.5\Omega)=60V$
$V_{12.5\Omega}=(4.0A)(12.5\Omega)=50V$
The required voltage of the battery is given as
$\epsilon=V_{15\Omega}+V_{12.5\Omega}+19.091V$
We plug in the known values to obtain:
$\epsilon=60V+50+19.091V=129V$
(b) We know that if a $17.2\Omega$ resistor is increased, then the resistance of the parallel sections of the circuit increases and as a result the current in the circuit decreases.
