Calculate the equivalent resistance of the network shown in the diagram. The resistances are: R1 = 12 ohm, R2 = 5 ohm, R3 = 15 ohm, and R4 = 15 ohm.
no prob here
Req = [(1/12)+(1/5)+(1/15)]^-1 + 15 = 17.86ohm
When a battery with E = 6 V and zero internal resistance is attached across this network, what current flows through resistor R1?
Cant figure out if the current thru R1 is actually Req or R1.
When i use V=IR none of the answers come out correct.