The circuit: http://www.webassign.net/walker/21-33alt.gif The question reads: Consider the group of resistors in the figure below, where R1 = 14.1 Ω and R2 = 8.35 Ω. The current flowing through the 8.35-Ω resistor is 1.22 A. What is the voltage of the battery? I found that the equivalent resistance was 15Ω + 14.1Ω + 4.74Ω = 33.84Ω, and V=IR, so V = 33.84Ω x 1.22A = 41.3V. I don't know where I'm going wrong.