Answer
a) $I_1=0.022A$ right
$I_2=0.124A$ left
$I_3=0.102A$ up
b) $I_1=0.023A$ right
$I_2=0.124A$ left
$I_3=0.101A$ up
Work Step by Step
a) $R_{eq}=R_3+\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}=53.3\Omega$
Lets assume the direction of the current is right for $R_1$, left for $R_2$, and up for $R_3$.
$I_1-I_2+I_3=0$
$I_1=I_2-I_3$
$I_1R_1+I_2R_2=9.0V$
$I_2R_2+I_3R_3=12.0V$
$I_3R_3-I_1R_1=3.0V$
$I_3R_3-(I_2-I_3)R_1=3.0V$
$I_3(R_3+R_1)-I_2R_1=3.0V$
$I_3=\frac{3.0V+I_2R_1}{R_3+R_1}$
$(I_2-I_3)R_1+I_2R_2=9.0V$
$I_2(R_1+R_2)-\frac{3.0V+I_2R_1}{R_3+R_1}R_1=9.0V$
$I_2(93\Omega)-\frac{3.0V+(25\Omega)I_2}{60\Omega}(25\Omega)=9.0V$
$I_2(4080\Omega)-75.0V+(875\Omega)I_2=540V$
$I_2=0.124A$
$I_3=\frac{3.0V+(0.124A)(25\Omega)}{60\Omega}=0.102A$
$I_1=0.124A-0.102A=0.0224A$
$I_1=0.0224A$ right
$I_2=0.124A$ left
$I_3=0.102A$ up
b) $I_1=I_2-I_3$
$I_1(R_1+r)+I_2R_2=9.0V$
$(I_2-I_3)(R_1+r)+I_2R_2=9.0V$
$I_2R_1+I_2r-I_3R_1-I_3r+I_2R_2=9.0V$
$I_3=\frac{I_2(R_1+r+R_2)-9.0V}{R_1+r}$
$I_3=\frac{I_2(94.0\Omega)-9.0V}{26.0\Omega}$
$I_3=3.62I_2-0.346A$
$I_2R_2+(3.62I_2-0.346A)(R_3+r)=12.0V$
$I_2(68\Omega)+(3.62I_2-0.346A)(36\Omega)=12.0V$
$I_2=\frac{24.5V}{198\Omega}=0.124A$
$I_3=3.62(0.124A)-0.346A=0.101A$
$I_1=I_2-I_3=0.124A-0.101A=0.023A$
