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

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To calculate the current through an 8Ω resistor connected to two 10V cells with 4Ω internal resistance, the total equivalent resistance must be considered. The initial calculation suggested a main current of 1.5A, but it was clarified that not all of this current flows through the 8Ω resistor due to the parallel configuration with the internal resistances. The correct approach involves recognizing that some current also flows through the internal resistance, leading to a final current of 1A through the 8Ω resistor. An alternative method involves simplifying the circuit by treating the two 4Ω resistances in parallel, resulting in an effective resistance of 10Ω, which confirms the 1A current using Ohm's law. Understanding the distribution of current is essential for accurately solving such circuit problems.
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Homework Statement


Ln4Uh.png

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

so is my concept right?
thanks a loto:)
 
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.
 
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