Electrical Engineering: Basic Laws (11 of 31) Kirchhoff's Laws: A Medium Example 2
Summary
TLDRThis video demonstrates how to apply Kirchhoff's laws to solve a two-loop circuit with a voltage source and three resistors. It illustrates using Kirchhoff's current and voltage laws to derive three equations, allowing calculation of currents through all resistors. The analysis starts by assuming current directions and applying Kirchhoff's first law at a node. Then, Kirchhoff's second law is applied to each loop to derive voltage equations. The resulting equations are solved simultaneously to find the currents. Although other methods may be easier, this approach effectively illustrates Kirchhoff's laws in action.
Takeaways
- 🔋 The circuit involves a two-loop configuration with a voltage source and three resistors.
- 🌀 Kirchoff's Laws are used to solve for the currents in all three resistors, providing a clear example of applying these rules.
- 🔄 The first assumption involves the direction of current flow through the three resistors, and if the assumptions are incorrect, negative values indicate the actual direction.
- 📏 Kirchoff's First Law (Current Law) states that the sum of currents entering a node must equal the sum of currents leaving that node.
- 🔁 A node is identified where current i1 enters, and currents i2 and i3 leave, leading to the first equation: i1 = i2 + i3.
- 🔋 Kirchoff's Second Law (Voltage Law) is applied to both loops, stating that the sum of voltage drops and rises around a loop must equal zero.
- 🔄 Loop 1 results in a voltage rise across the battery and voltage drops across the resistors, leading to the second equation.
- 🔁 Loop 2 also gives another equation, involving voltage rises and drops across the resistors, forming the third equation.
- 🔧 The system of equations is solved by substitution and elimination, ultimately calculating i2 and i3.
- 💡 The solution shows that i1 = 3A, i2 = 2A, and i3 = 1A, validating the node current splits and direction assumptions.
Q & A
What is the purpose of using Kirchhoff's laws in this circuit analysis?
-Kirchhoff's laws are used to solve for the current flowing through all three resistors by creating equations based on the sum of currents at a node and the sum of voltage drops in loops. This method is illustrated to show how to solve a circuit that could also be solved using simpler methods, like finding equivalent resistance.
What are Kirchhoff's two laws mentioned in the script?
-The first law (Kirchhoff's Current Law) states that the sum of currents entering a node must equal the sum of currents leaving the node. The second law (Kirchhoff's Voltage Law) states that the sum of voltage rises and drops around any closed loop must add up to zero.
How are the current directions assumed in this circuit?
-The current directions are assumed arbitrarily for each resistor. If the actual direction differs, the result will be a negative value, indicating that the assumed direction was incorrect. However, this won't affect the solution.
What is the first equation derived from Kirchhoff's Current Law (KCL)?
-The first equation derived from KCL is i1 = i2 + i3, where i1 is the current entering the node, and i2 and i3 are the currents leaving the node.
How are the two loops defined in the circuit?
-The two loops are defined by tracing a path around the circuit. Loop 1 is traced clockwise around the first part of the circuit, and Loop 2 is traced counterclockwise around the second part of the circuit. The sum of voltages in each loop must equal zero according to Kirchhoff's Voltage Law.
What is the second equation derived from Loop 1?
-The second equation from Loop 1 is 30 - 8i1 - 3i2 = 0. This represents the voltage rise from the 30V source and the voltage drops across the 8-ohm and 3-ohm resistors in the direction of the current.
What is the third equation derived from Loop 2?
-The third equation from Loop 2 is 3i2 - 6i3 = 0. This represents the voltage rise across the 3-ohm resistor (going against the current) and the voltage drop across the 6-ohm resistor (in the direction of the current).
How are the three equations solved simultaneously?
-To solve the system, i1 is substituted into the second equation to get two equations with i2 and i3. Then, the equations are manipulated to eliminate one variable, allowing for the solution of i2 and i3. Finally, i1 is found using the first equation (i1 = i2 + i3).
What are the final values of the currents i1, i2, and i3?
-The final values are i1 = 3A, i2 = 2A, and i3 = 1A. These values satisfy both Kirchhoff’s Current Law and Voltage Law, meaning the solution is correct.
What would happen if the directions of the currents were assumed incorrectly?
-If the current directions were assumed incorrectly, the final solution would give a negative value for the current, indicating that the actual current flows in the opposite direction to what was assumed. However, this doesn't affect the validity of the results.
Outlines
🔧 Introduction to Kirchoff’s Laws and Circuit Analysis
This paragraph introduces a two-loop circuit and outlines the process for solving it using Kirchoff’s laws. The circuit consists of a voltage source and three resistors. While other methods like finding equivalent resistance might be easier, the problem is being solved with Kirchoff’s laws for illustrative purposes. The goal is to determine the current through each resistor. The directions of the currents through each resistor are assumed, noting that a negative result would indicate the current is in the opposite direction. The paragraph also introduces the first of Kirchoff’s laws: the sum of currents entering a node must equal the sum of currents leaving the node, forming the basis for the first equation, i1 = i2 + i3.
🔄 Applying Kirchoff’s Voltage Law to Loop Analysis
In this section, the focus shifts to Kirchoff’s second law, which states that the sum of all voltages around a loop must be zero. Two loops are identified in the circuit, and the voltage rises and drops around each loop are analyzed. For Loop 1, the voltage rise from the battery is balanced by the voltage drops across resistors. The first loop equation is derived as 30 - 8 * i1 - 3 * i2 = 0. Similarly, in Loop 2, the second equation is derived by analyzing the voltage changes across the resistors, resulting in 3 * i2 - 6 * i3 = 0. With three equations now formulated (one from Kirchoff’s current law and two from voltage law), the groundwork is laid for solving the currents through the circuit.
Mindmap
Keywords
💡Kirchhoff's Laws
💡Node
💡Loop
💡Resistor
💡Voltage Source
💡Current
💡Voltage Drop
💡Simultaneous Equations
💡Assumed Current Direction
💡Voltage Rise
Highlights
Introduction to Kirchhoff's laws and their application to a two-loop circuit.
The circuit consists of a voltage source and three resistors: Resistor 1, Resistor 2, and Resistor 3.
Discussion on alternative methods like equivalent resistance for easier solving, but focus is on illustrating Kirchhoff's laws.
Assumption of current directions in the resistors, and how getting a negative result indicates the direction is opposite.
Kirchhoff's first law (Junction Rule): The sum of currents entering a node equals the sum of currents leaving the node.
Application of the Junction Rule at a specific node: i1 = i2 + i3.
Defining two loops for applying Kirchhoff's second law (Loop Rule) around the circuit.
Kirchhoff’s second law (Loop Rule): The sum of voltage rises and drops around a loop must equal zero.
First equation from Loop 1 using Kirchhoff's second law: 30 - 8*i1 - 3*i2 = 0.
Second equation from Loop 2 using Kirchhoff's second law: 3*i2 - 6*i3 = 0.
Combining the current equation (i1 = i2 + i3) into the loop equations to solve for currents i1, i2, and i3.
Solving the system of equations using substitution and elimination to find i2 = 2A, i3 = 1A, and i1 = 3A.
Verification of the solution: Currents in the branches (i2 = 2A, i3 = 1A) add up to the total current i1 = 3A.
Explanation of how this method demonstrates the use of Kirchhoff's laws, despite other simpler methods available.
Summary of the process: Using Kirchhoff's first and second laws to form three equations, then solving them simultaneously to find the current through each resistor.
Transcripts
welcome to alector line here we have
another example of how to apply Kirk's
laws here is a two-loop circuit we have
a voltage source we have three resistors
now we know that this problem can be
solved much easier probably using the
methodology of finding the equivalent
resistance and then finding the total
current and then see how the current
branches out in the various branches but
in order to illustrate how to use kirov
laws this is a good example to use here
we need to find the current through all
three resistors we have resistor one
resistor 2 and resistor 3 and one
voltage source we can see that through
the first resistor let's say that we
assume that there will be a current
flowing through this resistor in this
direction then here we have a current
flowing through this resistor let's
assume it's going to be in this
direction and here we have a current
flowing uh through this Branch right
here and through this resistor let's
call that I sub3 and again assume that
will be direct the direction of the
current now we could be wrong here it
could be that the direction of current
is in a different direction opposite of
what we drew but that's quite all right
if we did we'll end up with a negative
answer which that indicates the arrow
was drawn in the wrong direction
although with the voltage source having
the positive end on this side the
negative end on this side it can be
almost certain that the we can assume
that the current will be in this
direction as we have drawn the first law
of kirov states that we can add up all
all the currents entering a
node and that must equal all the
currents leaving that
node and if we use that law we can take
a look and see how about this node right
here let's pick this Noe we can see that
we have i1 entering that node and I2 and
I3 leaving that node based upon that we
have our first equation that i1 must
therefore equal I2 + I3 that's equation
number one
now we have two Loops we have a loop and
let's say that we go around the loop in
this direction call that Loop
one and the second Loop let's say we're
going to go around the loop in this
direction let's call that Loop
two we need to start at some node let's
say we started this node right here for
both loops and we'll go around this Loop
in this direction we'll go around this
Loop in this direction and the second
law of Kirk house says that the sum of
all the voltages around any Loop must
add up to zero which means when we add
up all the voltage Rises and all the
voltage drop around Loop one they should
add up to zero when when we add up all
the voltage Rises and voltage drops
around Loop two they should add up to
zero as well and that will give us the
other two equations we're looking for
because after all we're trying to solve
for three currents therefore we will
need three equations to solve this
problem starting at this node going
around the loop like this we go across a
30 volt battery that is a 30 volt rise
now we go across this resistor in the
same direction as the current that means
we have a voltage drop of minus the
resistance 8 times the current
i1 then we go to this uh note right here
through this resistor along the same
direction as the current again that's a
voltage drop so minus the Resistance 3
ohms times the current I2 now we have
made a complete Loop therefore this
should add up to zero and there's
equation number two
the second Loop starting again from here
we go across this resistor but now
notice we go against the current flow
therefore that's a voltage rise that' be
3 *
I2 here we go across this resistor in
the same direction of the current that's
a voltage drop minus the resistance 6
times the current I3 and now we've made
a complete Loop that must add up to
zero
notice we now have three equations one
two and three obtained by using both
laws of kirkov now we need to solve
those simultaneously the best way to do
that is to take the current equation
which is now already expressed as one
current in terms of the other two
currents and we can then substitute that
into the other two equations in this
case we only have to substitute it into
one equation because the second equation
does not include I sub one that will now
give us the following two equations
equation number two now changes to 30 -
8 * i1 and we know from this equation
that i1 is equal to I2 +
I3 minus 3
I2 equal to Zer the second equ or the
third equation now becomes well nothing
changes it is still 3 I2 - 6 I3 is equal
to
Z well we should do now is we need to
solve these two equations simultaneously
let's put them into a different format
we'll have the eyes on the left side and
any constants on the right side so for
the first equation the 30 will move to
the right side become a minus 30 we have
a minus 8 I2 minus 3 I2 that is a - 11
I2 - 8 I3 is equal to Aus 30 and since
all the terms in there are negative we
can simply multiply both sides by Nega 1
and and turn everything into a positive
so there's the equation number two in
the format with all the eyes on the left
side and the constant on the right side
now we can take the third equation
equation number three and write it as 3
I2 - 6 I3 is equal to
zero now we need to solve these
simultaneously usually the easiest way
is to go ahead and try to manipulate the
two equations multiplying times some
constant so that when we add the two
equations together we eliminate one of
the variables we notice that the that
the common uh if we multiply this by
three and this by four we we get 24 we
get a positive 24 a negative - 24 so
when we add the two equations the i3s
drop out so I'm going to multiply this
equation by three I'm going to multiply
this equation by four so the second and
third equation now become as follows
this equation becomes 33
I2 + 24 I 3 is equal to 90 the third
equation becomes 12 I2 - 24 I3 is equal
to zero if I now add the two equations
together notice the I3 will drop out I
end up with
45 I2 equal 90 and from this we can
conclude if we divide both sides by 45
we get I2 is equal to 2 and of course
that's 2 mamp might as well put the unit
in so now that we know that I2 is equal
to 2 Amp we can use this equation to
solve for
I3 move the minus 6 I3 to the other side
I end up with 3 I2 is equal to 6 I3
dividing both sides by 3 I get I2 is =
to 2 I3 or I3 is = to 12 I2 since I2 is
equal to 2 Amp that's 1 12 * 2 Amp other
words I3 is equal to 1 amp now I have
both I2 and I3 now I use the equation
the first equation with the currents to
come up with the value for i1 this is
equal to I2 which is 2 amps plus I3
which is 1 amp i1 is equal to 3
amps and that makes sense now when you
go over here and say okay if the current
through this branch is i1 equal to 3
amps and then the current splits up into
I2 and I3 if I2 is equal to 2 amps and
I3 is equal to 1 amp they do add up to 3
amps and that should be correct so
that's how we use Kirra rules to come up
with enough equation to solve for all
the currents in the circuit again this
circuit could be solved in many
different ways probably easier than this
way but it's a good illustration we
simply have a single current source and
some resistors we simply find a branch
point or a node we can add all the
current entering the node all the
currents leaving the node we set that
equal to each other for our first
equation then we go around two of the
loops Loop one Loop two add up all the
voltages around the loop add up all the
voltages that gives us equation two and
three and now we can solve those
equations simultaneously to find the
current in each branch and in each
resistor and that's how we do
that
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