Calculating Current Through 8Ω Resistor with 2 Cells: Homework Solution

[asdff529]

Homework Statement

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?

Homework Equations

V=IR

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

 

[gneill]

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

 

[asdff529]

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

so is my concept right?
thanks a lot

 

[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.

 

[Nickie]

I would like to first clarify that the calculation provided in the attempt at a solution is incorrect. The equivalent resistance of the two cells and the 8Ω resistor in series is not 4Ω, but rather 8Ω. Using Ohm's Law (V=IR), we can determine that the current through the 8Ω resistor is 1A, as stated in the correct answer. This can also be confirmed by using Kirchoff's Laws to calculate the total current in the circuit, which would also be 1A.

Additionally, it is important to note that the internal resistance of the cells will affect the overall current in the circuit. The higher the internal resistance, the lower the current will be. Therefore, it is important to take into account the internal resistance when calculating the current in a circuit.

Furthermore, the statement that "the current will separate at the point which joint the resistor and the cell" is not entirely accurate. The current will flow through the entire circuit, including the internal resistance of the cells, in a series circuit. It is only in a parallel circuit that the current will split at a junction.

In conclusion, to calculate the current through the 8Ω resistor in this circuit, we must take into account the internal resistance of the cells and use the correct equivalent resistance of 8Ω. This will result in a current of 1A, as stated in the correct answer.