# A problem on current

1. Jan 6, 2013

### asdff529

1. The problem statement, all variables and given/known data

two cell,with emf=10V,with 4Ωinternal resistance are connected with a 8Ω resistor as shown in the fig.
What is the current through the 8Ω resistor?

2. Relevant equations
V=IR

3. The attempt at a solution
If i just consider 1 cell but the internal resistance of another cell is remained there.I guess the current will separate at the point which joint the resistor and the cell.
Here is my calculation
Eq resistance=4+1/(1/8+1/4)
Main current=1.5A
Therefore the current passes through the resistor is 1A
Now come back to the problem,there are 2 cells,so i multiply 1 by 2 and get 2A
But the ans is 1A only

2. Jan 6, 2013

### Staff: Mentor

Check your current-divider calculation; The bulk of the current should be flowing through the smaller resistance of a parallel pair

3. Jan 6, 2013

### asdff529

so is my concept right?
thanks a lot

4. Jan 6, 2013

### CWatters

Your approach is correct but you didn't finish it. The figure of 1.5A is correct BUT not all of it flows through the 8 Ohm resistor. Some goes through the 8 Ohm and some goes through the 4 Ohm of the "shorted voltage source" that is in parallel with it. Calculate what fraction flows through the 8 Ohm and double that.

or you can cheat and take a different approach. You can connect any two points with the same voltage together without effecting the circuit. So with the aid of one wire you can reduce the circuit to 2 x 4 Ohm in parallel feeding an 8 Ohm = 10 Ohm. 10V/10 Ohms = 1A...but that wouldn't teach you about superposition.