Answer
a) $R_{eq}=\frac{41}{15}\times175\Omega=478\Omega$
b)$I_{top bottom}=0.1045A$
$I_{vertical}=0.0767A$
Work Step by Step
a) Here, we need to remember that if there are multiple paths the current can take at any node, it is in parallel and if there is only one path, it is in series.
Lets simplify the circuit by taking the 3 resistors on the far right. They are in series, so we can write this as one resistor with resistance 3R.
This is in parallel with another resistor so the total resistance is $\frac{1}{\frac{1}{R}+\frac{1}{3R}}=\frac{3}{4}R$
This is in series with two other resistors, so the total resistance is $2R+\frac{3}{4}R=\frac{11}{4}R$
This is in parallel with another resistor, so the total is $\frac{1}{\frac{4}{11R}+\frac{1}{R}}=\frac{11}{15}R$
Lastly, this is in. series with 2 resistors, so the sum is $\frac{11}{15}R+2R=\frac{41}{15}R$
$R_{eq}=\frac{41}{15}\times175\Omega=478\Omega$
b) The current drawn from the battery is $I=\frac{V_{AB}}{R_{eq}}=\frac{50.0V}{478\Omega}=0.1045A$
The top and bottom resistors are in series, so they have the same current running through them.
$V=IR=(0.1045A)(175\Omega)=18.3V$
$I_{top bottom}=\frac{V}{R}=\frac{18.3V}{175\Omega}=0.1045A$
Some of this current goes through the vertical resistor and the rest go through the rest of the resistors on the right. We know the voltage across the vertical resistor and the rest of the resister is the same.
$V_R=V_{\frac{11}{4}R}$
$I_{vertical}R=I_{2}\frac{11}{4}R$
$I_{2}=\frac{4}{11}I_{vertical}$
From Kirchoff's Law, we know
$I_{topbottom}=I_{vertical}+I_2$
$0.1045A=\frac{15}{11}I_2$
$I_{vertical}=0.0767A$
