Find the Voltage in This Circuit

[Dawg91]

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.

 

[SHISHKABOB]

well, the current in R2 won't be the current through the battery because of how the circuit's junctions are set up

try looking at that junction and remember the rules about voltages and currents in resistors that are in parallel

 

[gneill]

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.

The given current, 1.22A, flows through R2. It is not the entire current; there will be other currents in in other parts of the circuit. The equivalent resistance of the entire resistor network is going to be of limited use here. The two parallel resistors in the middle branch can be simplified, though, as can the series combination of R1 and 15.0Ω.

Start with the current that you're given and see what potentials you can deduce from that. Draw them in on your circuit diagram. As you go, see what other information becomes obtainable.