Been stumped on these 2 questions for at least a week. Both diagrams scanned + uploaded. 23.Find the current in the 8 ohm resistor. 1/Req1=1/8+1/4 2.67+6=8.67 1/Req2=1/8.67+1/10 R=4.27+.5+5=10.14 I first found the voltage drop in each resistor. I=V/R I=9/10.14=.89A volt.drop .5 resistor=.5*.89=.445V volt.drop 5 resistor=5*.89=4.45V volt.drop 4.27 resistor=4.27*.89=3.8V Then I solved for current in the 8 ohm resistor. I=[9V-(.89A)*(.5*5)]/x ohm And that's where I'm stuck. Don't know what ohm to divide by. I think 25 but I don't know why. 35.What would current I1 be if the 12 ohm resistor were shorted out(r=1 ohm)? OK...this is where I'm just screwed by Kirchhoff's Laws. I'm thinking: I3=I2+I1 Loop 1: 12=(1)I1+(8)I2 Loop 2: 12=(1)I2+(10)I2+(12)I2 Loop 3: 6=(1)I3+(18)I3+(15)I3+(10)I2+(1)I2 Completely stuck. Help please.