Electrical Engineering: Basic Laws (10 of 31) Kirchhoff's Laws: A Medium Example 1
Summary
TLDRThis 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
🔄 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.
🔋 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.
Mindmap
Keywords
💡Kirchhoff's Laws
💡Voltage Source
💡Resistor
💡Current Direction
💡Voltage Drop
💡Ohm's Law
💡Loop Equation
💡Assumed Current
💡Voltage Rise
💡Negative Voltage
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
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Kirchhoff's voltage law | Circuit analysis | Electrical engineering | Khan Academy
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