Electrical Engineering: Basic Laws (12 of 31) Kirchhoff's Laws: A Harder

Summary

This 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

00:00

🔋 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.

05:01

🔄 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.

Keywords

💡Kirchhoff's Rules

Kirchhoff's Rules refer to two fundamental laws used to analyze electrical circuits: Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL). These rules are central to the video's theme, helping solve for unknown currents in multi-loop circuits. The script discusses using Kirchhoff’s rules to determine current in a circuit with multiple branches and voltage sources.

💡Current

Current is the flow of electric charge through a conductor. In the video, the current in each branch of the circuit (I1, I2, I3) is analyzed. The direction of current flow is assumed initially, and corrected if the results are negative, indicating a reverse flow direction.

💡Voltage

Voltage refers to the electric potential difference between two points. It drives the current through a circuit. In the video, multiple voltage sources (batteries) are part of the circuit, and Kirchhoff's Voltage Law is applied to find the voltage drops and rises across different components.

💡Resistor

A resistor is a passive electrical component that resists the flow of current, causing a voltage drop. The video specifically mentions resistors with values of 4 ohms, 6 ohms, and 8 ohms, and these resistances are key to calculating the voltage drops and currents using Kirchhoff’s laws.

💡Loop

A loop in a circuit refers to any closed path where current can circulate. Kirchhoff’s Voltage Law is applied to loops to ensure that the sum of all voltage changes around any closed circuit loop equals zero. The video illustrates two loops, where equations are set up to analyze the voltage and current relationships.

💡Node

A node is a point in a circuit where two or more components are connected. Kirchhoff's Current Law is applied to nodes, stating that the total current entering a node equals the total current leaving it. The script uses this law to form the first equation that relates currents I1, I2, and I3 at a specific node in the circuit.

💡Assumed Direction of Current

When solving circuits using Kirchhoff’s laws, the direction of current is assumed and drawn. If the assumption is incorrect, the calculated current will come out negative, indicating the true direction is opposite. In the video, the instructor initially assumes the current directions, and later finds that I3 is negative, meaning the current flows in the opposite direction.

💡Voltage Drop

A voltage drop occurs when current flows through a resistor, resulting in a decrease in electrical potential. In the script, each loop's voltage drop is calculated using Ohm’s Law (V = IR), and the sum of these drops around the loop is set to zero, according to Kirchhoff’s Voltage Law.

💡Equation Substitution

Equation substitution is a method used to solve systems of equations, where one equation is substituted into another to reduce the number of variables. The video demonstrates this by substituting the expression for I2 (I1 + I3) into the other loop equations, simplifying the problem to find the values of I1, I2, and I3.

💡Negative Current

A negative current result means the initial assumption about the direction of current flow was incorrect. In the video, I3 comes out negative, indicating that the actual current flows in the opposite direction to the one initially assumed. This is a normal outcome in circuit analysis.

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