KVL question for multiple loops

AI Thread Summary
The discussion focuses on applying Kirchhoff's Voltage Law (KVL) to a circuit with multiple loops, specifically regarding the correct sign in the voltage equation for a right-hand loop. The user is uncertain whether the equation should be I2*R1 + I3*R2 + I3*R3 = 0 or -I2*R1 + I3*R2 + I3*R3 = 0, based on the direction of current entering resistor R1. A response clarifies that the sign depends on the chosen direction for KVL and the current's direction relative to that choice. It emphasizes that consistency in current direction and loop direction is key, and if they align, the terms are positive; if they oppose, they are negative. Understanding these principles ensures accurate application of KVL in circuit analysis.
Tsandman
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Hi, The picture below is part of the notes given to us by our teacher. In case it doesn't work, it's a double loop with a voltage source on the left arm, a resistor (R1) on the middle arm, and two resistors in series (R2 and R3) on the right arm. all the resistors and the voltage source are positive at the node at the top of the page.
The current is I1 in the left arm, I2 in the middle arm, and I3 in the right arm.

For summing the voltages around the right hand loop with a clockwise summing loop, is it:

I2*R1 + I3*R2 + I3*R3 = 0

or

- I2*R1 + I3*R2 + I3*R3 = 0

The teacher said it's the second one, but i thought that seeing as the summing loop enters R1 from the negative end the sign is different?

Thanks in advance for any help.

http://screencast.com/t/QTDiBUruA
 
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Hi Tsandman! :smile:
Tsandman said:
… i thought that seeing as the summing loop enters R1 from the negative end …

I don't know what you mean by this. :confused:

All that matters is the whether the arrow for the current is the same or opposite to the direction that you've chosen for KVL.

You can choose the other direction, in which case the whole equation gets multiplied by minus-one (so it still works).

Or you can choose to draw, say i2, the other way round (so you're using I2 = -i2), in which case only the coefficient of I2 will be multiplied by minus-one, and you multiply it by minus-one again when you convert it back to i2.

(btw, your i's are not the same as the i's in the diagram …)

Draw arrows for the currents, and a circular arrow for your loop …

if the arrows match, that's plus, if they don't, that's minus. Simples! o:)
 
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