Electrical Engineering: Basic Laws (10 of 31) Kirchhoff's Laws: A Medium Example 1

Summary

This video delves into an example of Kirchhoff's laws applied to a single-loop circuit. The instructor walks through solving for current and voltage across components, including resistors and voltage sources. The problem involves summing the voltages around the loop, accounting for drops and rises using assumed current directions. As the analysis unfolds, the calculations reveal a negative current, indicating that the actual current flows in the opposite direction of the initial assumption. By solving the equations, the voltage across one of the sources is found to be -48V and the actual current is 8 amps.

Takeaways

• 🔄 The example in the video focuses on using Kirchhoff's laws in a single-loop circuit with added complexity.
• 🔋 The circuit contains three voltage sources and two resistors, and Kirchhoff’s law states that the sum of voltages in the loop should add up to zero.
• 🔧 Even though the current is assumed to flow clockwise, it may actually flow in the opposite direction, which would affect the voltage drop across resistors.
• ⚡ A key equation used is V = IR to determine the relationship between current and voltage across the resistors.
• 🌀 The voltage drop across the 6-ohm resistor is calculated, taking into account that the assumed current direction may be incorrect.
• 🧮 Substituting values, the final equation is solved to find the current in the circuit.
• 📉 The current is calculated to be -8 amps, indicating that the actual current direction is counterclockwise instead of the assumed clockwise direction.
• 🔋 The voltage across the unknown voltage source is calculated as 48 volts using the current and resistance values.
• 🔄 The result is that the actual current in the circuit flows in the opposite direction to what was initially assumed.
• 📐 The key takeaway is the application of Kirchhoff's laws and Ohm's law to solve for unknown currents and voltages in a more challenging single-loop circuit.

Q & A

What is Kirchhoff's Voltage Law (KVL) and how is it applied in this circuit?

Kirchhoff's Voltage Law states that the sum of all voltages around any closed loop in a circuit must equal zero. In this example, the sum of the voltage rises and drops as you travel around the loop is calculated to ensure they total zero, helping solve for unknown values like current and voltage.

Why does the direction of the current not matter initially in this analysis?

The direction of the current doesn't matter because you can assume any direction for analysis. If the current ends up being negative, it means the actual direction is opposite to the assumed one. The sign of the current reveals the true direction after solving the equations.

What is the significance of voltage sources in this circuit?

There are three voltage sources in the circuit. One of them is unknown, but its voltage is defined in relation to the voltage drop across a resistor. These sources influence the total voltage sum as you traverse the loop.

How is the voltage drop across a resistor calculated?

The voltage drop across a resistor is calculated using Ohm's law, which states that the voltage drop equals the resistance multiplied by the current (V = IR). If you cross a resistor in the direction of the current, it results in a voltage drop, hence a negative term in the KVL equation.

How does the script handle the relationship between the unknown voltage and the resistor?

The script states that the voltage across the unknown source is equal to twice the voltage drop across a 6-ohm resistor. This is incorporated into the equations to help solve for both the current and the unknown voltage.

Why do they initially assume the current is clockwise, and what happens when this assumption is incorrect?

The current is initially assumed to be clockwise for simplicity. If the calculation shows a negative current value, it indicates that the actual current flows in the opposite direction.

How is Ohm's law used to relate the current and the unknown voltage?

Ohm's law (I = V/R) is applied to the voltage drop across the 6-ohm resistor. By substituting this into the overall KVL equation, the script solves for both the current and the voltage.

What final equation is used to solve for the current in the circuit?

The final equation used to solve for the current is 12 - 4I + 12I + 4 - 6I = 0, which simplifies to 2I = -16, leading to I = -8 amps.

Why is the current value negative, and what does it indicate about the direction of current flow?

The current value is negative because the initial assumption about the current direction was incorrect. This means the actual current flows counterclockwise, not clockwise as initially assumed.

What is the voltage across the unknown voltage source, and how is it calculated?

The voltage across the unknown voltage source is calculated as V = -6 * (-8) = 48 volts. Since the current direction is opposite to the assumed one, the final voltage value is positive.

Outlines

00:00

🔄 Understanding Kirchhoff's Laws in a Circuit Loop

In this paragraph, the speaker introduces a slightly more challenging example of Kirchhoff's laws applied to a single-loop circuit. The goal is to sum up all the voltages around the loop in a clockwise direction, following Kirchhoff's law that the sum of all voltages in a closed loop must equal zero. Three voltage sources and two resistors are noted, including a voltage source with an unknown value, which is equal to twice the voltage drop across a 6-ohm resistor. Despite the assumed clockwise current direction, the voltage drop across one of the resistors suggests the current might actually be flowing in the opposite direction. The current direction doesn't impact the solution, as any resulting negative value indicates the actual current flow.

05:03

🔋 Summing Voltages Around the Circuit

Here, the speaker begins calculating the voltages around the circuit, starting from point A. Moving across different components, the speaker notes a 12-volt rise across a battery, a voltage drop across a resistor (calculated as minus four times the current), and additional voltage changes across other circuit elements. They acknowledge that even though some voltage drops suggest the actual current may differ from the assumed direction, this won't change the final outcome. The goal is to add up the voltages and create an equation that will help solve for the current and voltage in the circuit.

📉 Using Ohm's Law to Find Voltage and Current

The paragraph details how Ohm's Law (I = V/R) is used to establish a relationship between current and voltage. The voltage drop across a 6-ohm resistor is calculated, but the assumed direction of the current creates a situation where the actual voltage behavior differs from initial assumptions. After re-calculating with the correct polarity and direction of voltage rise, the speaker derives the current (I) and integrates this into the earlier voltage equation, simplifying it to solve for the unknown current and voltage in the circuit.

🔢 Solving the Equations for Current

At this point, the speaker solves the system of equations derived from Kirchhoff's and Ohm's laws. After simplifying the terms, they find the current (I) to be -8 amps, meaning the initial assumption of current direction was incorrect. A negative value indicates that the actual current is flowing in the opposite direction. The solution provides clarity on how incorrect assumptions regarding direction can be corrected once the current is calculated.

⚡ Calculating Voltage Across the Unknown Voltage Source

Finally, the voltage across the unknown source (V₀) is calculated using the previously determined current. With the current value of -8 amps, V₀ is found to be 48 volts. This result confirms the earlier calculation, showing that the voltage across the source is negative, indicating a voltage drop from left to right. The speaker concludes by explaining the actual direction of the current (positive in counterclockwise or negative in clockwise) based on the assumptions made during the problem-solving process.

Keywords

💡Kirchhoff's Laws

Kirchhoff's laws are fundamental principles in circuit analysis, specifically the laws of voltage and current. Kirchhoff's Voltage Law (KVL) states that the sum of all voltages around a loop equals zero, while Kirchhoff's Current Law (KCL) states that the total current entering a junction equals the total current leaving it. In the video, KVL is applied to sum up the voltages around a single loop in the circuit.

💡Voltage Source

A voltage source is a component in a circuit that provides electrical energy. It can be a battery or another source that maintains a fixed voltage. In this circuit, there are three voltage sources, including one unknown source whose voltage is defined in relation to the resistor’s voltage drop. Understanding how these sources affect the circuit is key to applying Kirchhoff's laws.

💡Resistor

A resistor is a component that resists the flow of electric current, creating a voltage drop proportional to the current passing through it. Ohm's Law (V = IR) governs this relationship. The circuit in the video includes two resistors, one of 6 ohms and another of 4 ohms, which play a crucial role in determining the current flow and voltage distribution across the circuit.

💡Current Direction

The direction of current flow is assumed when analyzing a circuit. In this example, the current is initially assumed to flow clockwise, but the video shows that the actual current is in the opposite direction based on the negative sign in the calculated current. The assumption of current direction can be corrected later if the results show a different actual flow.

💡Voltage Drop

A voltage drop occurs when electrical energy is converted to heat or another form of energy in a resistor. In the video, the voltage drop across each resistor is calculated based on the assumed current direction. These voltage drops are subtracted from the total voltage sources to ensure Kirchhoff's Voltage Law holds true in the loop.

💡Ohm's Law

Ohm's Law relates the current flowing through a conductor to the voltage and resistance, expressed as V = IR. It is used in the video to determine the relationship between the voltage across a resistor and the current through it. The law is applied to calculate the voltage drops across the resistors in the circuit.

💡Loop Equation

A loop equation is formed by summing all the voltage rises and drops around a closed circuit loop, as mandated by Kirchhoff's Voltage Law. In the video, the loop equation is written by starting at point A, summing voltage contributions from batteries and resistors, and setting the total equal to zero.

💡Assumed Current

Assumed current refers to the initial guess of the direction of current flow in the circuit. In the video, the current is assumed to be clockwise, but the final calculation reveals the current actually flows counterclockwise. The negative sign in the current result indicates the original assumption was incorrect.

💡Voltage Rise

A voltage rise occurs when moving from a lower to a higher potential, typically when crossing a battery from its negative to positive terminal. In the video, a voltage rise is encountered when crossing a 12-volt battery, contributing positively to the loop equation. This concept contrasts with voltage drops across resistors.

💡Negative Voltage

Negative voltage refers to a voltage source or voltage drop that is in the opposite direction to the assumed reference point or current flow. In the video, the voltage across the unknown voltage source (V₀) is calculated to be -48 volts, indicating that it opposes the assumed direction of current flow. This negative sign helps clarify the true behavior of the circuit.

Highlights

Introduction to a more challenging example of Kirchhoff's laws, focusing on a single-loop circuit with multiple voltage sources and resistors.

Explanation of Kirchhoff's Voltage Law (KVL): The sum of all voltages around any loop in the circuit must always add up to zero.

Observation that the current is assumed to flow clockwise, but the actual current might flow in the opposite direction, which will be determined by the sign of the result.

The first voltage rise is encountered across a 12-volt battery as we move from the negative to the positive end.

The voltage drop across the 4-ohm resistor is calculated as minus 4 times the current (I), following the assumed direction of current flow.

The voltage drop across an unknown voltage source is described as minus 2 times V₀ (the unknown voltage).

A 4-volt rise is observed across another voltage source in the loop.

The voltage drop across the 6-ohm resistor is described as minus 6 times the current (I), completing the voltage summation around the loop.

Kirchhoff's law equation is formulated, combining all the voltage rises and drops, and is set to equal zero.

Ohm's law (I = V/R) is used to relate the current and voltage across the 6-ohm resistor, where V₀ is determined to be equal to -6I.

Substituting V₀ = -6I into the Kirchhoff's law equation allows for solving the current in the circuit.

The current in the circuit is calculated to be -8 amps, confirming that the actual current flows in the opposite direction to the assumed one.

The voltage across the unknown voltage source (V₀) is found to be 48 volts, with a negative sign indicating a voltage drop.

The negative current value reflects the opposition to the assumed clockwise current direction, indicating that the true current flows counterclockwise.

Final conclusions: the current in the circuit is 8 amps counterclockwise, and the voltage across the unknown source is -48 volts.

Transcripts

welcome to electron line in this video

we're going to take another look at an

example of Kirchhoff's laws this one is

going to be a little bit more

challenging for us it is still a single

loop example and what we're going to do

here is travel around the loop in a

clockwise direction add up all the

voltages of course using kirchoff's laws

we know that they should add up to zero

we know that's the sum of all the

voltages around any loop in the circuit

must always add up to zero let's start

at a and go around the circuit notice

that there are three voltage sources

notice there's two resistors this

voltage source is not known but we know

that the voltage across this source is

equal to twice the voltage drop across a

six ohm resistor also notice that even

though we're assuming the current is in

a clockwise direction the voltage drop

across this resistor would indicate that

the actual current is in the opposite

direction but again that shouldn't

really matter we can choose the

direction any way we like and if the

answer then comes out to be negative we

know that the assumed direction is

actually the incorrect direction and the

real current is flowing in the opposite

direction all right let's go ahead and

sum up all the voltages going from A to

B we cross the 12-volt battery from the

negative end to the positive end that's

a 12-volt rise now we go across the

resistor for ohms times the current

there would be a minus four times I

voltage drop whenever we cross a

resistor in the direction of the assumed

current we assume a voltage drop here we

have a voltage drop across this voltage

source equal to minus 2v sub naught

minus 2v sub naught now we have a

voltage rise across this battery or this

voltage source o plus four and finally

even though we know that the actual

voltage drop is from the negative into

the positive in this direction we're

just going to assume the direction the

current is clockwise and therefore we

assume a voltage drop six ohms times the

current minus six times the current and

that adds up to zero a second equation

that we can use to solve the problem

because what we're trying to do here is

solve for the current in the circuit and

solve for the voltage

across this voltage source we're going

to use Ohm's law I equals V over R to

determine a relationship between the

current and VSIP not using what's

happening on this resistor here we can

see that the voltage drop across this

resistor is VSIP naught but since we

assume the current to be in this

direction that would cause this to be a

voltage drop and since action looks like

it's a voltage rise this therefore is

equal to minus 6 V sub nought that's the

voltage drop across the resistor it's in

essence a voltage rise divided by the

resistance so I'll hold it hold it hold

it back up so this is equal to the

voltage across that it is going to be a

minus V sub nought why - because we're

assuming the current to be in this

direction which causes which should make

this a voltage drop but instead it's a

voltage rise so we have to put minus in

there divided by the resistance which is

6 ohms solving this for V sub naught we

can then see that V sub naught across

the resistor is equal to minus 6 times I

which we then substitute into our

equation here so we need to plug that

right in there that allows us to solve

for the current 12 minus 4 times I minus

2 times V sub naught which is a minus 6i

plus 4 minus 6 I is equal to 0 let's go

ahead and get rid of parentheses 12

minus 4i plus 12 I plus 4 minus 6i is

equal to 0 adding up all the currents on

the left side minus 4 plus 12 minus 6

that is 2i and moving all the voltages

or the numbers to the other side that

becomes a minus 12 and 8 minus 4 or 2 I

is e to the minus 16 or I is equal to

minus 8 amps dividing both sides by a 2

now we know the current in the circuit

and notice a sign and negative 8 amps we

assume that that would be the case that

we would have a real current in the

opposite direction because of the

voltage drop that was in opposition to

what we assumed when we have the current

going in this direction now we can solve

for the voltage across the voltage

source V sub naught is equal to minus 6

times I and I is a minus 8 amps of

course this is a minus 6 ohms let's just

put minus 6 ohms here and ohms times

amps is your voltage so therefore V sub

naught is equal to 48 volts so the

solution is such that voltage across

this battery is a minus 48 volts notice

from left to right the voltage drops

across the source and here we know how

babba voltage rise because we know the

actual direction of the current I actual

is equal to 8 amps in whoa I'll take

that back the current is actually in

this direction I actual is equal to a

positive 8 amps if we assume a counter

clockwise direction and it's some

negative 8 amps if we assume a clockwise

direction and those are the answers for

this particular problem