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