Transcript Go With the Flow

Alex Kearns
Richard Kerr
Rocio Rodriguez
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An electric circuit is a collection of electrical
devices connected by conductors, usually wires, to
a power source, which supplies a flow of electrons.
Electricity is defined as this flow of electrons
originating from and returning to the power
source.
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Voltage: potential difference in charge between two points in an electrical
circuit
represented by the uppercase letter V
Unit of voltage is the Volt (joule/coulomb)
Potential difference is directly proportional to the force that pushes
electrons/current from one point to the other
 Electrical potential difference can be thought of as the ability to move electrical
charge through a resistance
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Resistance: The opposition that a substance offers to the flow of electric
current.
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represented by the uppercase letter R
unit of resistance is the ohm (volt*meter)
Current: A flow of electrical charge carriers, usually electrons.
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represented by the uppercase letter I
unit of current is the ampere (coulomb/second)
one ampere of current represents one coulomb of electrical charge (6.24 x 1018
charge carriers) moving past a specific point in one second.
Any voltage V, current I or resistance R in an
electrical circuit can be determined without
actually measuring it if the two others values
are known.
 This law can be used to determine the amount
of current I flowing in the circuit when voltage
V is applied to resistance R. Ohm's law is:
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Current = Voltage / Resistance.
In the following circuit, assume that resistance
R is 2 and voltage V that is applied to it is 12 V.
Then, current I flowing in the circuit can be
determined as follows:
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Series Circuit:
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An open in the circuit will disable the entire
circuit.
The voltage divides (shared) between the
loads.
The current flow is the same throughout the
circuit.
The resistance of each electrical device can be
different.
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The total resistance RO in this circuit is equal to the
sum of individual resistance R1 and R2.
In other words: The total resistance(RO) is equal to the
sum of all resistances (R1 + R2 + R3 + .......)
Therefore, the strength of current (I) flowing in the
circuit can be found as follows:
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Parallel Circuit:
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A Parallel Circuit has multiple paths or
branches to ground. Therefore:
In the event of an open in the circuit in one
of the branches, current will continue to
flow through the remaining.
Each branch receives source voltage.
Current flow through each branch can be
different.
The resistance of each branch can be
different.
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Resistance R0 (a combination of resistances R1 and R2)
in a parallel connection can be determined as follows:
From the above, the total current I flowing in this
circuit can be determined from Ohm's law as follows:
The total current I is also equal to the sum of currents
I1 and I2 flowing through individual resistances R1
and R2
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Kirchhoff’s Node Law:
the sum of all the currents entering
a node is equal to the sum of all the
currents leaving the node.
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Kirchhoff’s Voltage Law:
The directed algebraic sum of the voltages and
voltage drops around a loop must be zero.
In other words, the sum of the voltage drops
will always equal the voltage provided by the
power source.
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Assign currents to each part of the circuit
between the node points. We have two node
points Which will give us three different
currents. Lets assume that the currents are in
clockwise direction.
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So the current on the segment EFAB is I1, on the segment BCDE is
I3 and on the segment EB is I2
Using the Kirchhoff's current Law for the node B yields the
equation
I1 + I2 = I3. For the node E we will get the same equation.
Then we use Kirchhoff's voltage law
-4 I1+ (-30) -5 I1 - 10I1 +60 +10I2 =0
When through the battery from (-) to (+), on the segment
EF, potential difference is -30, and on segment FA moving
through the resistor of 5W
will result in the potential difference of -5 I1 and in a similar way
we can find the potential differences on the other segment of the
loop EFAB.
In the loop BCDE, Kirchhoff's voltage law will yield the
following equation:
-30 I3+ 120-10I2+60 =0
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Now we have three equations with three unknowns:
I1
+ I2 - I3
=0
-19 I1 +10 I2
= - 30
-10 I2 -30 I3
= -180
The system above has the following solution:
I1 =2.8302
I2 =2.3774
I3 =5.2075
This linear system could have been solved using linear
algebra with much less complicated math. Linear
Algebra is more useful when the network is very
complicated and the number of the unknowns is large.
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Write down Kirchhoff’s law for each loop. The result,
after simplification, is a system of n linear equations in
the n unknown loop currents in this form:
where R11, R12, . . . , Rmm and V1, V2, . . . , Vm are
constants.
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Solve the system of equations for the m loop currents I1,
I2, . . . , Im using Gaussian Elimination.
The number of loop currents required is 3.
Let’s choose the loop currents shown.
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Write down Kirchhoff's Voltage Law for each
loop. The result is the following system of
equations:
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Put these equations in a matrix and solve using
Gaussian Elimination:
Solving the system of equations using Gaussian
elimination gives:
I1 = - 4.57, I2 = 13.7 and I3 = - 1.05
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