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Hi Folks,
Got a question on network analysis thats been really puzzling me would appreciate any help on this. In a circuit containg two DC voltage sources and three resistors, if E1 is 28 V and E2 is 7 V and there are three resistors in the circuit R1(4 ohms) is in series with E1. R3 (1 ohm) is in series with E2. R2(2 ohms) is a common branch in the middle of the circuit. If one of the tereoms are used to calculate the currents flowing through all resitors the answers are: I1= 5A, I2 = -1A and I3 = 4 A.
However because I2 is negative it is considered an error and the actual current is flowing in the oppposite direction however if that is true then KVL does not work.
With I2 negative the volt drops are:
Loop 1 (VR1= 20 V, VR2 = 8)  and loop 2 ( VR3 -1, VR2 =8V)
This to me is correct because this agrees with KVL.
However if the current is consisdered positive as it should be then loop 2 = 9 V which does not agree with KVL.
I Hope I have explained my problem clearly again any help will be appreciated.

Cheers JAg.

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A circuit sketch would help but I think you are confusing current flow with polarity.
While the polarity of the current does indicate the direction of the flow, the magnitude
Of the current is always a positive number.  3Amps * 1V = 3watts.  -3amps * 1v = 3watts
Not Negative 3 watts.  A negative wattage would indicate that heat is being sucked away from something.  Even the famed pettler junction doesn’t really do that as the other side still gets very hot.

I assume the method you are using is to remove 2 batteries from the circuit and then calculate all the currents for the entire circuit with just one battery, then repeat the process for the other two batteries, and finally sum all the results?

This will show you the actual current flows and voltages in the circuit.  It is tedious but it works.

Good luck
-Mike

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In the first picture I've drawn a classical representaion of the voltages across the elements in a simple battery/load circuit. I am careful to observe that the voltage arrows always point to the higher potential end of the component. When this is so, the voltage is always marked against the arrow as positive. We have thus eliminated any ambiguity regarding polarity, prior to applying Kirchoff's laws.

To apply Kirchoff's voltage law we must add all the voltages in a loop, without forgetting to account for polarites. In the left example, adding +10V to +10V is 20V, not zero, giving the impression that something is wrong. Nothing is really wrong here, of course, except that clockwise arrows all must be added, and anti-clockwise all subtracted (or vice versa if you wish) in order to correctly account for the various polarities. So in this example, the battery voltage (clockwise arrow) is positive, and the load R's arrow is anti-clockwise, and thus negative. Add those two: plus10 + minus10 = 0.

In the right version I made a visual tweak to the load's arrow - I changed it's direction, and negated the voltage (to maintain polarity consistency). This visual trick now means all the arrows point clockwise, and can all be added as-is. The sum is zero, as you can see.

For your own circuit, I worked out the layout and values, and drew in arrows and values paying careful attention to polarity. The arrow "VR2" points to the right-hand end of R2, and is marked "-1V". Read this to mean "The right-hand end of R2 is minus 1V higher than the left-hand end". In other words, the right end is one volt lower.

You could change the arrow direction, and label it +1V, and this would say exactly the same thing.

When observing consistency in your labelling this way, Kirchoff's voltage law is always applicable. Using the existing labels, as I have drawn them, we see that VR3 and VR2 point clockwise, and so are added, but E2 is anti-clockwise, so is subtracted. So:

VR3 + VR2 - E2 = (+8)+(-1)-(+7) = 0

There is another interesting observation to make regarding your circuit, and that is the direction of current through E2. Take another look at my simple battery/load circuit, and note how all the labelled values are positive. In this condition, all arrows are correctly depicting direction of current flow and polarity of potential difference, and can be used to visually spot odd circumstances.

See how the current flow through the resistive load and voltage drop across it point in opposite directions? This is normal for a resistive load, and indicates that the load is absorbing energy. In other words, the resistor is getting warmer.

What if (given all-positive current and voltage labels) the current direction and potential difference are in the same direction? That means that the component in question is delivering energy! In my picture that is exactly the case for the battery, and is perfectly normal considering batteries are power sources.

But in your circuit there is an unusual situation for the battery E2. The current flowing through it is I2, which is labelled (at the moment) as -1A. First, normalise that arrow by changing it's direction and sign, so that we are using only positive values. Now you can see that for the battery E2 the two arrows (I2 and E2) point in opposite directions, telling you that this battery is absorbing energy. In other words it is being charged up! This battery would therefore heat up and possibly explode if it was not rechargeable!  ## Join the conversation

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