# Find current in Resistor 2

1. Feb 23, 2012

### Cisneros778

1. The problem statement, all variables and given/known data
I do not understand the solution provided for this problem.

Consider the circuit with five resistors and two batteries (with no internal resistance) shown in the figure.

(b) Solve for the current in the 4-Ω resistor.

2. Relevant equations
loop law
junction law

3. The attempt at a solution

The answer below is correct, however I am confused on how equations (2) and (3) were found.

(1) 12 - i1 -3i2 =0
(2) 6 -5 i1 +8i2 -4i3 =0
(3) 6 + 2 i1 -2 i2 -6i3 =0

solving the above we get
i1= 4.16
i2 = 2.61
i3 =1.51

current through 4-O resistor = i3 = 1.51 A
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 23, 2012

I think you are confused between KVL and KCL that is loop law and junction law resp. as you mentioned. No the answer is not right and since its loop equation given as sum of all the voltages in a loop but you are using junction law with it.

3. Feb 24, 2012

### Redbelly98

Staff Emeritus
It looks like (2) is the loop equation for the loop 6V-3Ω-4Ω-5Ω, and (3) is the loop equation for the triangular loop in the upper right part of the circuit.

Try writing out these loop equations yourself. To do that you'll need to figure out:
The current in the 5Ω resistor, in terms of i1 and i2, and
the current in the 2Ω resistor, in terms of i1, i2, and i3.​