Answer
(a) $1.44A$
(b) $11.64V$
(c) decrease
Work Step by Step
(a) We can find the required current as follows:
$R_{eq1}=(\frac{1}{3.2\Omega}+\frac{1}{7.1\Omega+5.8\Omega})^{-1}=2.564\Omega$
$R_{eq}=0.25\Omega+4.5\Omega+1.0\Omega+2.564\Omega=8.31\Omega$
Now $I=\frac{V}{R_{eq}}$
$I=\frac{12V}{8.31\Omega}$
$I=1.44A$
(b) We know that
$V=\epsilon-Ir$
We plug in the known values to obtain:
$V=12V-(1.44A)(0.25\Omega)$
$V=11.64V$
(c) We know that
$R_{eq1}=2.564\Omega$. If the resistance $3.2\Omega$ increases, the equivalent resistance $R_{eq1}$ increases and as a result the equivalent resistance of the entire circuit $R_{eq}$ will increase and thus, the current $I=\frac{V}{R_{eq}}$ will decrease.
