Answer
$I_{1,2}=0.284mA$
$I_3=0.852mA$
$I_4=2.27mA$
$I_5=0.568mA$
$I_6=1.42mA$
$V_{AB}=1.85V$
Resistor naming. Starting from the battery and going clockwise, name each resistor on the outer path 1 to 4. Then, name the resistor between A and B 5 and the resistor between A and C 6.
Work Step by Step
First, let's name all resistors. Starting from the battery and going clockwise, name each resistor on the outer path 1 to 4. Then, name the resistor between A and B 5 and the resistor between A and C 6.
First, let's find the equivalent resistance.
series(1 & 2): $R_1+R_2=2R$
parallel( & 5): $\frac{1}{\frac{1}{2R}+\frac{1}{R}}=\frac{2}{3}R$
series( & 3): $\frac{2}{3}R+R=\frac{5}{3}R$
parallel( & 6): $\frac{1}{\frac{3}{5R}+\frac{1}{R}}=\frac{5}{8}R$
series( & 4): $\frac{5}{8}R+R=\frac{13}{8}R=5280\Omega$
$I_4=\frac{V}{R_{eq}}=\frac{12.0V}{5280\Omega}=2.27mA$
$I_4=I_6+I_3$
$V_6=V_3=12.0V-(2.27mA)(3250\Omega)=4.62V$
$I_6=\frac{4.62V}{3250\Omega}=1.42mA$
$I_3=2.27mA-1.42mA=0.852mA$
$I_3=I_{1,2}+I_5$
$V_{1,2}=V_5=4.62V-(0.852mA)(3250\Omega)=1.85V$
$I_5=\frac{1.85V}{3250\Omega}=0.568mA$
$I_{1,2}=0.852mA-0.568mA=0.284mA$
$V_{AB}=(I_5)(R)=(0.568mA)(3250\Omega)=1.85V$
