Batteries Series Parallel Circuit

How do I find the current in the resistance of a series-parallel circuit?

How much current flows through each lamp in the figure. P35? Lamp 1 has a resistance of 3.0 ohms, Lamp 2 has a resistance of 5.0 ohms and lamp 3 has a resistance of 7.0 ohms. The battery has a resistance Internal negligible. Please http://www.webassign.net/hecht/18-P35alt.gif studies show: D

To find total current, find strength equivalent. Req = 3 * 5 / (3 +5) + 7 = 8.875 ohm current in the lamp 3 – 6V / 8.875ohms = 0.68 in the lamp current 2 A – 3 / (3 +5) * .68 A = 0.255 current through the lamp 1 A – 5 / (3 +5) * .68 A = 0.425 A

Physics I J1_ Batteries in Series and Parallel (ΔV)-Bob Abe