Circuit with Resistors and Two Opposing EMFs

AI Thread Summary
The circuit analysis involves two batteries with opposing EMFs and resistors, leading to a complex current flow. The top battery has an EMF of 16V and the bottom 8V, with resistances of 5.0Ω, 9.0Ω, and internal resistances of 1.6Ω and 1.4Ω. The calculated current magnitude is 8/17 A, but the direction is confirmed to be counterclockwise due to the stronger top EMF. A key point in circuit analysis is that if the calculated current is negative, it indicates the actual flow is opposite to the assumed direction. Proper application of Kirchhoff's voltage law (KVL) is essential for determining current direction and magnitude accurately.
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Homework Statement


The circuit shown in Fig. 25.37 contains two batteries, each with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit (magnitude and direction) . . .

The circuit is a rectangle with the emfs at the top and bottom sides and the resistors on the left and right sides. The positive side of each emf is towards the left side, so that one creates a current clockwise and the other an opposing one counterclockwise. The resistances are: left, 5.0; right, 9.0; top (internal), 1.6; bottom (internal), 1.4. The emf at the top is 16 and at the bottom 8.

Homework Equations


V = IR
V = E - Ir
I = E / (r+R)

The Attempt at a Solution


For the formula I = E / (r+R), since the denominator is the sum of all the resistances, I assumed I could sum all the emfs as well (taking sign and direction into account). So I did:

I = (16-8) / (1.4+5+1.6+9) = 8/17

I do not know what the correct answer is, but I intuitively feel like this answer is wrong.
 
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The current magnitude is correct. How is the current drawn on the circuit, is it flowing clockwise or counterclockwise?
 
Wow, so that is the correct magnitude? That's good. The direction should be counterclockwise, since the top emf is stronger than the bottom one, and the top one is going counterclockwise.

And just for clarity I drew a rough sketch of the diagram:

PhysicsCircuit.png
 
Yes, your magnitude and direction are correct. Just remember that when doing a KVL around a loop, if your answer is positive your assumed current direction is correct. If negative, the direction of the current is opposite of the assumed direction.


For example, say I was summing the voltages around the loop in the clockwise direction:

16V + I(9) + (-8V) + I(1.4) + I(5) + I(1.6) = 0V
I = -8/17 A

Since I assumed a clockwise direction for the current, the answer came out negative. This is telling me the current is actually going in the opposite of the assumed direction.
 
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