Circuits, Currents, and Kirschoff’s Laws

A PROJECT BY JAMES SABO AND SALLY JUNE TRACY

VOLTAGEELECTRIC POTENTIAL ENERGY DIFFERENCE

PER UNIT CHARGE BETWEEN TWO POINTS

VOLTS (joules/coulomb)

CURRENTMOVING ELECTRIC CHARGE IS IN THE DIRECTION OF POSITIVE

CHARGE FLOW.

NET RATE AT WHICH CHARGE CROSSES A GIVEN AREA

DRIVEN BY VOLTAGE DIFFRENCE

AMPERES (coulombs/second)

RESISTANCE• In most conductors charge doesn’t

move unimpeded • Charge looses energy gained from

electric potential difference in collisions

• This is described by the quantity of resistance specific to different conductors

• In units of Ohms (volt meters)

V = IR

• The relationship between voltage current and resistance

• The potential drop across a circuit element is linearly proportional to the current

R1

I1

V = IR in Practice

• Lets suppose V1 = 6V and R1 = 3Ώ

• V = IR I = V/R

• I = (6V/ 3Ώ) = 2A

Kirschoff Loop Law

• The sum of the voltages in a circuit is equal to zero

• Across a resistor:• With the current becomes

negative (voltage drop)• Against the current

becomes positive (voltage gain)

• V1 - I1R1 - I1R2 = 0• Or • I1R1 + I1R2 = V1

V1

R1

R2

Kirschoff Node Law

• The sum of the currents into and out of a node are equal to zero.

• Current in is positive• Current out is negative• I1 – I2 – I3 = 0 I1 = I2 + I3

Kirschoff’s Laws in Action

• As you can see, with multiple voltage sources and multiple paths, V = IR will not solve for the currents here.

• Kirschoff’s laws must be applied.

A

Apply Node Law

• At point A: I1 + I2 - I3 = 0

• Or I1 + I2 = I3

• If the current direction is not known, simply guess.

• Based on the circuit, it looks like a good guess for the direction of each, but if we are wrong, don’t worry! The current will just become negative.

A

Apply Loop Law

• You can pick up to three different loops here. The two shown and one could be picked around the perimeter.

• From loop 1:

• V1 - I1R1 - I3R2 = 0

• From loop 2:

• V2 - I2R3 - I3R2 = 0

Loop

1

Loop

2

Solving the Circuit

• We have three equations and three unknowns, the circuit can now be solved.

• WARNING, This section involves rigorous algebra, you may become lost, confused or bored in the process.

• Just bear with us it will be over soon enough.

Solving the Circuit

• From the three equations:• 1. I1 + I2 = I3 I1 = I3 – I2

• 2. V1 - I1R1 - I3R2 = 0

• 3. V2 - I2R3 - I3R2 = 0

• Sub 1. Into 2.

• V1 – (I3 – I2)R1 - I3R2 = 0• Solve for I3.

• 4. I3 = (V1 - I2R1)/(R1 + R2)

• Sub 4 into 3.

• V2 - I2R3 - [(V1 - I2R1)/(R1 + R2)] R2 = 0

• Solve for I2

• I2 =[V2 – (V1R2)/(R1 + R2)] / [(R3 – (R1R2)/(R1 + R2)]

• Use given values:• I2 = 1.20A

Still Solving the Circuit

• Plug I2 into 4.• I3 = (V1 - I2R1)/(R1 + R2)• I3 = 1.08A• Sub I2 and I3 into 1• I1 = I3 – I2

• I1 = -.13A• Now the circuit is

solved!

• As you can see that method involves a lot of algebra, and a lot of keeping track of which equation is where.

• Fortunately for us, there is a better way of doing things.

Linear Systems…My Savior!

• You have seen before how currents can be solved using rigorous algebra, but a much simpler method can be used.

• Fortunately we are certain that anyone in this class can solve this circuit using basic row operations we have learned in this class.

Use Linear Systems!

• Go back to the three original equations.

• 1. I1 + I2 = I3

• 2. V1 - I1R1 - I3R2 = 0• 3. V2 - I2R3 - I3R2 = 0• Put all currents on the

left and all non currents on the right.

• I1 + I2 - I3 = 0

• I1R1 + I3R2 = V1 • I2R3 + I3R2 = V2

• As you should be able to see, this can become a matrix.

• [ I1 + I2 - I3 : 0 ]• [ I1R1 + I3R2 : V1]• [ I2R3 + I3R2: V2]

Taking the RREF

• An augmented matrix can be formed.

• [ 1 1 -1 : 0 ]

• [ R1 0 R2 : V1]

• [ 0 R3 R2 : V2 ]

• Plug in the values

• [ 1 1 -1 : 0 ]

• [ 3 0 5 : 5 ]

• [ 0 8 5 :15 ]

• Find the RREF of the matrix

• [ 1 0 0 : -.1265822]• [ 0 1 0 : 1.202531 ]• [ 0 0 1 : 1.075949 ]• This corresponds to each

of the currents.• I1 = -.1265822A• I2 = 1.202531A• I3 = 1.075949A

The Mesh Method

• A different form of Kirschoff’s loop laws to solve for currents is called the Mesh Method.

• Instead of assigning a current through each resistor, assign a current through each loop.

• Once the currents are found, add the currents flowing through each resistor.

Mesh Method Application

• Going back to the old circuit we used, instead of using three equations (two loop laws and a node law) It can be broken into two equations.

• V1 - I1R1 - (I1 + I2)R2 = 0• V2 - I2R3 - (I2 + I1)R2 = 0

Loop

1

Loop

2

Mesh Method Application

• The currents can then be solved for in each equation and put into a matrix.

• I1(R1 + R2) + I2R2 = V1

• I1R2 + I2(R3 + R2) = V2

• [ (R1 + R2) R2 : V1 ]

• [ R2 (R3 + R2) : V2 ]

Mesh Method Application

• Plugging in original values, currents can be solved. The currents through each resistor are then added to solve for each current.

• [ (3Ώ + 5Ώ) 5Ώ : 5V ] [ 8 5 : 5 ]

• [ 5Ώ (8Ώ + 5Ώ) :15V] [ 5 13 : 15]

• RREF = [ 1 0 : -.1265822] I1 = -.1265822A

• [ 0 1 : 1.202531 ] I2 = 1.202531A

Solving for the Currents

• We now know that I1 = -.1265822A and that I2 = 1.202531A

• Therefore the current through R1 is I1= -.13A The current through R3 is I2 = 1.20A and the current through R2 is (I1 + I2) = 1.07A

Why use the Mesh Method?

• You may wonder why the mesh method can be helpful. In the last exercise, instead of 3 equations, there were 2. That doesn’t save too much time.

• When the circuits start to get more tricky, the mesh method starts to look a heck of a lot more useful.

Monster CircuitSix currents, which means six equations!

Quickly Analyze the circuit

• V3 - (I1 +I3)R3 - I1R6 - (I1-I2)R1 - V1 = 0

• V1 - (I2 - I1)R1 - I2R4 - (I2 + I3)R2 - V2 = 0

• V3 - (I1 + I3)R3 - I3R5 - (I2 + I3)R2 - V2 = 0

• I1(-R3 - R6 - R1) + I2(R1) + I3(-R3) =V1 - V3

• I1(R1) + I2(-R1 - R4 - R2) +I3(-R2) = V2 - V3

• I1(-R3) + I2(-R2) + I3(-R3 - R5 - R2) = V2 - V3

Put it in a matrix and take the RREF

• [-R3 - R6 - R1 R1 -R3 : V1 - V3 ]

• [ R1 -R1 - R4 - R2 -R2 : V2 - V1 ]

• [ -R3 -R2 -R3 - R5 - R2 : V2 - V3 ]

• Note that: • i1 = I2 - I1; i2 = -I2 - I3; i3 = I1 + I3

• i4 = I2; i5 = I3; i6 = I1 + I3

• AND WE’RE DONE!

Thank you for paying attention!

• We hope you all have a wonderful day and that you have somewhat been enlightened

in the ways of linear systems!

• Goodbye