Electrical Engineering: Basic Laws (12 of 31) Kirchhoff's Laws: A Harder
Summary
TLDRThis video explains how to solve a circuit problem using Kirchhoff's rules, particularly when multiple loops and voltage sources are involved. The speaker walks through determining the currents in various branches of the circuit, assuming initial directions for the currents, and formulating equations based on Kirchhoff's current and voltage laws. The process involves solving three equations to find the unknown currents, and adjusting for any negative results indicating opposite directions. The video concludes with calculated values for the currents in each branch, demonstrating the effective use of Kirchhoff's rules.
Takeaways
- 🔋 Kirchhoff's rules are ideal for analyzing circuits with multiple loops and voltage sources.
- 🔄 Assume a current direction in each branch; incorrect assumptions will lead to a negative result, indicating the opposite direction.
- ⚡ Kirchhoff's Current Law states that the sum of currents entering a node equals the sum of currents leaving the node.
- 🔁 Kirchhoff's Voltage Law says that the sum of all voltages around any closed loop should equal zero.
- 🌀 The circuit consists of three loops with currents I1, I2, and I3 flowing through 4Ω, 6Ω, and 8Ω resistors.
- 📐 For loop 1, the voltage equation is based on a 10V battery and resistors, resulting in: I1 + I3 = I2.
- ➕ For loop 2, the equation involves voltage drops and rises across the resistors and batteries, giving: 8I1 + 6I3 = -4.
- ➖ When simplifying and solving, the current I3 turns out to be negative, meaning the direction is opposite to the initial assumption.
- 🔍 Solving for I2 and I1 using substitution and combining equations results in positive values for these currents.
- ✅ Final results: I1 = 1.039 A, I2 = 0.731 A, I3 = -0.308 A, confirming the analysis using Kirchhoff's laws.
Q & A
What are Kirchhoff's rules, and why are they useful for solving this type of circuit problem?
-Kirchhoff's rules consist of two laws: Kirchhoff's current law (KCL) states that the sum of currents entering a node equals the sum of currents leaving the node. Kirchhoff's voltage law (KVL) states that the sum of all voltages around any closed loop in a circuit is zero. They are useful for solving circuits with multiple loops and voltage sources, like the one in this problem, where determining individual branch currents can be complex.
How are the current directions initially assumed in this circuit?
-The current directions are assumed based on the orientation of the voltage sources. For the branch with the 10V battery, current is assumed to flow clockwise (I1). For the branch with the 4V battery, current is assumed to flow clockwise (I3). If these assumptions are incorrect, the final calculated current values will simply be negative, indicating that the actual current flows in the opposite direction.
What are the main steps to solve for the currents using Kirchhoff's rules in this problem?
-1. Assume the direction of current in each branch (I1, I2, and I3). 2. Apply Kirchhoff's current law at a node to get the relationship between the currents. 3. Apply Kirchhoff's voltage law to two independent loops to form two more equations. 4. Solve the system of three equations to find I1, I2, and I3.
How is Kirchhoff's current law applied in this circuit?
-Kirchhoff's current law is applied at the top node, where the currents I1 and I3 enter the node, and I2 leaves the node. This gives the equation I1 + I3 = I2.
How is Kirchhoff's voltage law applied to loop 1?
-For loop 1, Kirchhoff's voltage law is applied by summing the voltage changes around the loop: starting at the 10V battery (positive), moving through the 4-ohm resistor (voltage drop of -4I1), and then through the 8-ohm resistor (voltage drop of -8I2). The total sum must equal zero, forming the equation 10 - 4I1 - 8I2 = 0.
How is Kirchhoff's voltage law applied to loop 2?
-For loop 2, Kirchhoff's voltage law is applied by summing the voltage changes: starting with the 8-ohm resistor (voltage rise of +8I2), moving through the 6-ohm resistor (voltage rise of +6I3), and then across the 4V battery (voltage drop of -4V). This gives the equation 8I2 + 6I3 - 4 = 0.
What is the significance of a negative current value after solving the equations?
-A negative current value indicates that the assumed direction for that current is opposite to the actual direction. For instance, if I3 is negative, it means the current flows in the opposite direction to what was initially assumed.
How do the equations change after substituting I2 in terms of I1 and I3?
-After substituting I2 = I1 + I3 into the voltage equations, the new equations become: 10 - 4I1 - 8(I1 + I3) = 0 and 8(I1 + I3) + 6I3 - 4 = 0. These are simplified to solve for I1 and I3.
What is the final value of current I3, and what does its negative sign imply?
-The final value of I3 is -0.308 amps. The negative sign implies that the actual direction of the current is opposite to the assumed clockwise direction in loop 2.
What are the final values of the currents I1 and I2?
-The final value of I1 is 1.039 amps, and the final value of I2 is 0.731 amps. Both currents flow in the assumed directions.
Outlines
🔋 Introduction to Kirchhoff's Rules and Circuit Analysis
This paragraph introduces the use of Kirchhoff's rules in solving complex circuits with multiple loops and voltage sources. The narrator highlights the goal of the video: to determine the current through the 4-ohm, 6-ohm, and 8-ohm resistors. The direction of current flow is assumed based on the battery's polarity, with the understanding that even if the assumption is incorrect, the result will simply indicate a negative value, showing the current flows in the opposite direction.
🔄 Kirchhoff's Laws and Loop Equations
Here, Kirchhoff's two laws are explained: (1) the sum of currents entering a node equals the sum of currents leaving the node, and (2) the sum of voltages around any loop must be zero. The speaker outlines the method of analyzing two loops, labeling them and defining the direction of travel around each loop. The goal is to generate three equations to solve for the unknown currents I1, I2, and I3 using these laws.
Mindmap
Keywords
💡Kirchhoff's Rules
💡Current
💡Voltage
💡Resistor
💡Loop
💡Node
💡Assumed Direction of Current
💡Voltage Drop
💡Equation Substitution
💡Negative Current
Highlights
Introduction to Kirchhoff's rules in a circuit with multiple loops and voltage sources, ideal for applying Kirchhoff's laws.
Explanation of the assumption process for current directions in each branch, showing that if the direction is wrong, the final answer will come out negative.
Kirchhoff’s current law is used: the sum of currents entering a node equals the sum of currents leaving the node.
Identification of two loops in the circuit and the importance of indicating the direction in which you follow each loop.
First equation is derived from the current law: I1 + I3 = I2.
Second equation is derived from the first loop using Kirchhoff's voltage law: sum of voltage drops around the loop equals zero.
Third equation is derived from the second loop, again using Kirchhoff's voltage law.
The second equation is substituted with I1 + I3 in place of I2 to simplify the system of equations.
The third equation is similarly updated by substituting I1 + I3 for I2.
Both simplified equations are set up for elimination to solve for I3, showing step-by-step calculations.
I3 is solved to be -0.308 amps, indicating that the initial assumption of current direction was incorrect for that branch.
I2 is solved to be 0.731 amps, confirming the assumed current direction in this branch is correct.
I1 is found to be 1.039 amps using the relationship I1 = I2 - I3.
The negative current value for I3 indicates that its actual direction is opposite to the assumed direction.
Conclusion: Kirchhoff's laws are effective in solving complex multi-loop circuits, even when initial assumptions about current direction are incorrect.
Transcripts
welcome to electron line now in this
video we have a more typical example of
how to use kirchoff's rules matter of
fact in this type of circuit kirchoff's
rules is ideal and the reason for that
is we have multiple loops and we have
multiple voltage sources that's usually
a good indication that you want to use
at least try kirchoff's rules on this
type of problem we're supposed to find
the current in each of the branches we
want to find the current to the 4 ohm
resistor to the 6 ohm resistor and to
the 8 ohm resistor how do we do that
well we start out by assuming a
direction in each of the branches
since the positive end of the battery is
on this side I'm going to assume that
the current flows in this direction
through this branch and I'll call this I
1 here we have the voltage source with
the positive end on this direction and
on this side so I'm going to assume that
the current flows in this direction
through this branch so let's call that I
sub 3 and then through this branch if
the two currents come together we can
assume that the current will flow in
this direction through here let's call
that I sub 2 now I may be wrong it could
be that I sub 3 is actually in the
opposite direction that the voltage but
the potential difference across this
battery is so large that it overwhelms
this one and since current in this
direction that is possible but it
doesn't matter if we assume the wrong
direction in the end if we assume the
wrong direction for I 3 the answer will
come out negative which then indicates
its acts in the opposite direction so
you don't have to sit there and worry oh
did I get that right even if you get it
wrong it doesn't matter now we have the
two laws of kerkhof the first one says
that the sum of all the currents
entering a branch or entering a node
should equal all the currents or the sum
of all the currents leaving the node and
also know that the sum of all the
voltages around in a loop should add up
to zero we're going to use two loops
here's loop number one it's always a
good idea to indicate the direction that
you're taking when you're going around
the loop in here that's loop number two
and again indicate the direction you're
going to follow when you go around loop
number two so let's come up with the
three equations we need three equations
because
yawns I 1 I 2 and I 3 if I take this
note right here I notice that I 1 plus I
3 add up to I 2 these are the two
currents entering the note and one
current leaving the note so equation
number one tells us that I 1 plus I 3
equals I to the second equation can be
found by going around loop number one I
can start at this note right here go
around look number one get to the same
note and add up all the voltages so
equation number two is obtained by going
from here across the battery from
negative to positive that's a positive
10 volts going across the resistor in
the same direction as the current means
we have a voltage drop so we get minus 4
times i1 coming down through this branch
here in the same direction the current
that's another voltage drop minus 8
times i2 then I get back to the same
point where I started so therefore this
should add to zero that's equation
number two equation number three can be
found by starting at any node in the
loop let's start with this node going
from here in a clockwise direction
across this resistor against the current
that means that the voltage rise that's
plus eight times the current i2 then in
this direction that's against the
current that the voltage rise plus 6
times I 3 and then from here to there
that's from the positive and the
negative end of the battery that's a
minus 4 volt drop hope we don't have to
write the unit's to save or and since we
can then get back to the same point that
adds up to 0 there's the three equations
and we can use those to find the three
unknowns I 1 I 2 and I 3 I'd like to use
this equation first because it usually
already has one of the currents in terms
of the other two we can take that to
substitute that into this equation right
here and into this equation right there
so the two equations two and three will
now change to the following two is now
going to be equal to 10 minus 4 times I
1 my
is eight times instead of writing I 2 we
can write I 1 plus I 3 I 1 plus PI 3 the
third equation same thing instead of
writing I 2 I can write PI 1 plus I 3
this gives me 8 I 1 plus I 3 plus 6i 3
is equal to not equal yet minus 4 is
equal to 0 simplifying the two equations
writing all the eyes on the left side
and all the constants on the right side
equation number 2 so we keep these two
going equation number 2 now becomes
minus 4 I 1 minus Zeta 1 is minus 12 I 1
minus a times I 3 is minus 8 I 3 equals
when we bring the 10 to the other side
becomes a minus 10 the third equation we
have 8 times I 1 8 times I 3 plus 6
that's plus 14 PI 3 and that equals when
we bring the minus 4 to the other side
becomes a positive 4 now notice that if
I multiply the second equation by 2 and
I multiply the third equation by 2 by 3
the minus 12 becomes the minus 24 the 8
becomes a plus 24 when I then have to do
equations together the I ones drop out
and only am left with a single unknown
which I can solve for so next the two
equations now become as follows I'm
multiplying this equation by 2
I'm multiplying this equation by 3 and I
get nope this is equation number 2
equation number 3 minus 24 I 1 minus 16
i 3 equals minus 20 in this equation
plus 24 PI 1 plus that would be 42 I 3
equals plus 12 when I add the two
equations together notice the I ones
drop out 42 minus 16 that would be 26 I
3 see 26 plus 16 that's 42 and this
becomes 20 minus 20 plus 12 minus 8
which means that I 3 is equal to minus 8
divided by 26 and let me get a
calculator for that 8 divided by 26
equals 0.3 0.9 Asst 0.30
8 amps notice that we end up with a
negative negative current which means my
initial assumption that the current was
in a counterclockwise direction in this
loop it's actually in a clockwise
direction so the negative indicates that
the actual current is in the opposite
direction but that's okay we can leave
it like that that we can then realize
the current is simply in this direction
now we need to solve for I 2 let's use
this equation right here to solve for I
2 8 times I to so I'm taking this
equation over here
8 times I 2 plus 6 times I 3 which is a
negative zero point 3 0 8 minus 4 equals
0 which means that 8 I 2 is equal to
positive 4 plus 6 times zero point 308
and of course we divide both sides by 8
by 8 that which means that I 2 is equal
to so multiplied times 6 plus 4 and
divided by 8 and I get point seven three
one zero point seven three one amps
which is the current in I 2 so I - I do
have the right the correct direction
because I got a positive answer finally
we can solve for I 1 using this equation
I can now say that I 1 is equal to I 2
minus I 3 so I 2 is equal to 0.73
1 amps - a minus 0.30 8 and that makes
that a positive this is therefore equal
to one point
zero three nine amps
there's I won it gives me the three
currents i1 i2 and i3 the way I've drawn
it
we now realize it's in the opposite
direction and that's how we use
Kirchhoff's laws to solve a multi Loop
multi voltage source problem like this
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