Applying Kirchoff's Voltage and Current Rules

RolandIRL

hi all,
i'm doing revision for physics and i'm having trouble with this particular problem. it's to do with kirchoff's voltage and current rules.

In the attachment is a circuit diagram for it. i've drawn in the currents as i think they should be but i think they might be wrong.
i applied KVL to the upper loop and the bottom loop. i then applied KCL to the middle junction on the right.
UPPER LOOP: V2 + I2R2 + I3R3 = 0 (1)
BOTTOM LOOP: V2 + I2R2 + I1R1 = V1 (2)
KCL: I1 = I2 + I3 (3)

Subtracting 1 from 2 i got,
V1 = I1R1 - I3R3

given V1,V2,R1,R2,R3, i still cannot find the current flowing through R2 (ie I2)
any help would be grateful.

Michael Collins

Hi whiteman19,

You can solve those equations by eliminating I2 from equations (1) and (2) by using equation (3).*

Then, given that you know the voltages V1 and V2, and all the resistances, you'll have two equations with two unknows (I1 and I3) and you can solve accordingly.

*But you may not want to do that as they arn't quite correct for the circuit you've given. You have to be very careful about sign convention when writing circuit equations. When current flows through a resistor, the voltage on the side it flows into is always more positive. For example when I2 flows into R2 in your picture, the right side of R2 is more positive and so the voltage across it has opposite sign to the source V2, but you have both voltages V2 and I2*R2 (the voltage dropped across R2) positive - a contradiction!

Try writing the equations out again keeping this in mind...I'll check them out if you want to post them up here.

RolandIRL

new circuit diagram drawn (units omitted for simplicity's sake) and sample values given to V1,V2,R1,R2,R3, just to solve the eqns.

Applying KVL to ABCF:
V2 + I2R2 + I3R3 = 0 ....(1)

Applying KVL to CDEF:
V2 + I2R2 = V2 +I1R1 .....(2)

Applying KCL to junction C:
I1 + I2 = I3 ....(3)

Subbing in values for voltages and resistances:
2*I2 + I3 = -1 (eq 1)
2*I2 - 3 *I3 = -3 (eq 2)

putting I2 = I3 - I1
eq 1 and 2 become
(1) 3*I3 - 2I1 = -1
(2) -2*I3 + 5*I1 = 3

solving these simultaneous eqns, i got I1 = 7/11 amps, I3 = 1/11 amps
plugging into (3) above, I2 = -6/11 amps (the minus indicates it flows in the opposite direction, doesn't it)
can someone verify this please?

Excuse the bad drawing :P

RolandIRL

could someone check this for me please? if i know how to do this one, i think i'll be able for all of problems like this. thanks

Michael Collins

Hey,

Those answers arn't quite right. But they are close.

Check the Right Hand Side of equation 2 (I presume you meant V1 and not V2 there) in particular check the sign of the voltage.

The right answers, including signs, are:

I1 = 5/11 A
I2 = -8/11 A
I3 = 13/11 A

and yes, the minus sign means you intially choose the incorrect direction, but that's fine.

RolandIRL

thanks, but what did you mean by the voltage signs? and it was meant to be V1

Applying KVL to ABCF:
V2 + I2R2 + I3R3 = 0 ....(1)

Applying KVL to CDEF:
V2 + I2R2 = V2 +I1R1 .....(2)

Applying KCL to junction C:
I1 + I2 = I3 ....(3)

(1) is right, i take it.
for (2) i put voltage drops on the left and voltage rises on the right. wait, should I1R1 be a voltage drop?
V2 - with the current flowing against the battery, V2 is a drop.
R2 and R1 are also voltage drops and V1 is a voltage rise. have i got this right? could you explain which are V drops and rises and why please?
thanks

Michael Collins

whiteman19 wrote: »

(1) is right, i take it.

Yep that's fine.

for (2) i put voltage drops on the left and voltage rises on the right.

It's all fine except for V1, which should be -V1 in your equation.

could you explain which are V drops and rises and why please?

When you're deciding if they voltage drops you do two things

1) put signs on all resistors with + on the side current is flowing in. You've done that perfectly, so no worries here
2) go around each loop in your circuit, writing the loop equation as you go, being consistant with the direction in which you meet signs.

I'll do this for the bottom loop in your circuit, which is the one you're having trouble with.

Starting at V1 and going clockwise:

First we meet V1, which is a voltage drop, so we start writing the equation:

-V1

next is R1, but with your chosen direction for I1, and going clockwise, this is a voltage rise. The equation becomes

-V1 + I1*R1

next we come across R2, which is seen to cause a voltage drop, so:

-V1 + I1*R1 - I2*R2

Finially, V2 is also a voltage drop:

-V1 + I1*R1 - I2*R2 - V2 = 0

or

V2 + I2*R2 = -V1 + I1*R1

Quite close to what you had, only for the sign of V1.

Does this make sense?

RolandIRL

i get it now. voltage drops is when you go high potential (+) to low potential (-) and voltage rises - vice versa. i think i was following the current direction in the lower loop and when i came to CF, iwas doing it wrong. you don't really take into consideration current when you do V drops and V rises, just the potentials on the batteries and resistors (although they were determined by the current) and so if a current is negative, the wrong direction was chosen. it all comes out in the algebra
i get it now. thanks

FISMA

Whiteman,

Kirchoff's loop rule is just another version of the Law of Conservation of Energy (like Bernoulli, FLoT, wave eq...) . It is very handy.

What you are doing is solving systems of equations" getting more eq's than unknowns.

Kirchoff's junction rule for i's is just conserving charge. The junction rule is a very quick and easy way to get an equation. Just choose a direction, who cares if it is right or wrong (for now). This is especially true when you deal with multiple battery circuits.

In general when you go across the battery from - to + that's a V rise. However, the battery could be getting charged, which could throw you off in a multiple EMF circuit.

Anyhow, do the math and solve for the currents and all. If your current turns out to be negative, than the direction you chose for that current was wrong.

It is a lot like the sum of the torques - if a torque turns out to be negative - you originally chose the wrong direction. Don't go back and change your work, just note it now...

As I am sure you probably already know, not all circuits break down to either series, para, or a combo thereof. That's when Kirchoff has to be used.