1. Consider the group of resistors in the figure, where R1 = 12.3 and R2 = 7.95[PLAIN]http://www.webassign.net/images/omegacap.gif. [Broken] The current flowing through the 7.95http://www.webassign.net/images/omegacap.gif resistor is 1.22 A. What is the voltage of the batter? http://i.imgur.com/zB2eulo.png 2. Series - Req=R1+R2+..., Parallel - 1/Req=1/R1+1/R2+..., Ohm's Law - V=IR 3. I started off by getting the equivalent resistance of the whole circuit, which I did by using the parallel equation for 13.8 and 17.2, which gave me 7.657 as the resistance of those two. Then I added 7.95 and 4.11 because those two are in series, which gave me 12.06. I believe the 12.06 and the 7.657 are now in parallel, which means that 1/12.06+1/7.657=1/Req. This gives me 4.68, which I believe is in series with the 12.3 and the 15 ohm resistors, so adding them all together gives me 31.98. I then take that 31.98 and plug it in to Ohm's law with the 1.22 A of current, which gives me 39.01 V. This is incorrect and I'm not sure where I went wrong. The obvious places to look would be making sure that each resistor was actually in series or in parallel with the one that I placed it with. I also am unsure if when they say that 1.22 A of current is flowing through the 7.95 resistor, if that means that the current for the entire circuit is 1.22 or if it has decreased by splitting up into separate currents and I have to calculate that. I have tried this several times and cannot figure out what I'm doing wrong. Thank you.