Practice Problem: Solving a Circuit Using Ohm's Law and Kirchhoff's Current Law
Summary
TLDRThis video script from the 12th edition of Irvine's textbook, problem 2.12, guides through a circuit analysis involving resistors and current dividers. The objective is to determine the currents I1 and I2, and the power absorbed by a 40 kiloohm resistor in a parallel circuit with a current source. By applying Kirchhoff's current law and Ohm's law, the script demonstrates the calculation of voltage (VX), current values, and power, concluding with I1 at 12 milliamps, I2 as -4 milliamps, and the power at 5.76 watts. The explanation emphasizes understanding circuit behavior, current direction, and the relationship between resistor values and current magnitude.
Takeaways
- 📚 The problem is from the 12th edition of 'Irvine' and involves resistors and current dividers.
- 🔍 The circuit has a current source in the middle with two resistors on each side, arranged in parallel.
- 👉 The task is to find the currents I1 and I2, and the power absorbed by the 40 kiloohm resistor.
- 🔄 Kirchhoff's current law is applied to find the relationship between the currents at the node.
- Ohm's law, represented as I = V/R, is used to relate voltage, current, and resistance.
- 🔌 The voltage across both resistors (VX) is the same because they are in parallel.
- ⚙️ The currents I1 and IB are calculated using the derived equations from Ohm's law.
- 🔢 The calculated voltage VX across the resistors is found to be 480 volts.
- ⚡ The current I1 through the 40 kiloohm resistor is 12 milliamps, and IB through the 120 kiloohm resistor is 4 milliamps.
- 🔁 Since I2 is in the opposite direction to IB, I2 is -4 milliamps.
- 💡 The power absorbed by the 40 kiloohm resistor is calculated to be 5.76 watts using P = IV.
Q & A
What is the main topic of the video script?
-The main topic of the video script is the analysis of a circuit involving resistors and current dividers, specifically focusing on finding the currents I1 and I2 and the power absorbed by a 40 kiloohm resistor.
Which textbook edition is the problem from?
-The problem is from the 12th edition of 'Irvine'.
What are the two main quantities the video aims to find?
-The two main quantities the video aims to find are the currents I1 and I2 in the circuit.
What is the significance of redefining the current I2 as IB?
-Redefining the current I2 as IB helps to simplify the analysis by allowing the current to be considered in the same direction for both resistors, which is necessary because they are in parallel and thus have the same voltage across them.
What law is applied to find the relationship between the currents in the circuit?
-Kirchhoff's Current Law is applied to find the relationship between the currents in the circuit.
How is Ohm's law used in the script to find the currents I1 and IB?
-Ohm's law, which states V = IR, is rearranged to I = V/R to find the currents I1 and IB, using the voltage VX across each resistor.
What is the calculated voltage VX across each resistor?
-The calculated voltage VX across each resistor is 480 volts.
What is the calculated current I1 flowing through the 40 kiloohm resistor?
-The calculated current I1 flowing through the 40 kiloohm resistor is 12 milliamps.
What is the calculated current IB flowing through the 120 kiloohm resistor?
-The calculated current IB flowing through the 120 kiloohm resistor is 4 milliamps.
What is the power absorbed by the 40 kiloohm resistor?
-The power absorbed by the 40 kiloohm resistor is 5.76 watts.
What insight can be gained from the direction of the currents I1 and I2 in the circuit?
-The insight gained is that even though I2 was initially defined in the opposite direction, redefining it as IB going in the same direction as I1 helps understand that in a parallel circuit, the current through each resistor can be different, and the actual direction of I2 is opposite to the initial arrow notation.
Outlines
🔌 Circuit Analysis with Resistors and Current Dividers
The video script begins with an introduction to a practice problem from 'Irving's 12th Edition E 2.12' that involves resistors and current dividers. The task is to find the currents I1 and I2 and the power absorbed by a 40 kiloohm resistor in a parallel circuit with a current source. The script describes the circuit setup, where two resistors are in parallel, and current directions are labeled. It emphasizes the need to understand the direction of current flow and power absorption by resistors. The script then introduces the concept of redefining current (IB) for simplicity and applies Kirchhoff's Current Law to set up an equation for the currents. Ohm's law is also discussed to relate voltage, current, and resistance.
🔍 Solving for Voltage and Current in a Parallel Circuit
This paragraph continues the circuit analysis by solving for the unknown voltage VX across the resistors using Kirchhoff's Current Law. The script provides a step-by-step calculation that leads to the determination of VX as 480 volts. It then uses Ohm's law to find the currents I1 and IB through the respective resistors. The calculations reveal that I1 is 12 milliamps and IB is 4 milliamps, with I2 being the negative of IB, indicating a current of -4 milliamps. The paragraph concludes with the calculation of the power absorbed by the 40 kiloohm resistor using the formula P = IV, resulting in 5.76 watts. The summary highlights the relationship between the resistor values, current magnitudes, and the direction of current flow, which is crucial for understanding the circuit's behavior.
📚 Applying Ohm's Law and Kirchhoff's Laws to Circuit Analysis
The final paragraph wraps up the problem-solving process by reiterating the application of Ohm's law and Kirchhoff's current law to solve the circuit. It emphasizes the importance of correctly identifying the direction of current flow and voltage across components in a parallel circuit. The script concludes with a brief discussion on the implications of the calculated currents and power, noting the relationship between the resistor sizes and the current distribution. It also touches on the concept of redefining current direction for clarity in circuit analysis, which is essential for a comprehensive understanding of the problem at hand.
Mindmap
Keywords
💡Resistors
💡Current Dividers
💡Kirchhoff's Current Law
💡Ohm's Law
💡Parallel Circuit
💡Voltage (VX)
💡Current (I1, I2, IB)
💡Power Absorption
💡Kilohms
💡Milliamps
Highlights
Introduction to a practice problem involving resistors and current dividers from the 12th edition textbook.
Objective to find currents I1 and I2 and the power absorbed by a 40 kiloohm resistor.
Description of the circuit with a current source and two resistors in parallel.
Identification of current directions and labeling for I1 and I2.
Understanding that resistors absorb power and cannot supply it.
Introduction of a new variable IB, defined as the negative of I2 for simplicity.
Application of Kirchhoff's current law at a node in the circuit.
Equation setup using Kirchhoff's current law: 16 milliamps equals I1 plus IB.
Use of Ohm's law to express current in terms of voltage and resistance.
Rewriting Ohm's law to solve for I1 and IB with given resistances and voltage VX.
Solving for VX using the equation derived from Kirchhoff's law.
Calculation of I1 and IB using the found voltage VX and resistance values.
Determination of I1 as 12 milliamps and IB as 4 milliamps, with I2 being the negative of IB.
Calculation of power absorbed by the 40 kiloohm resistor using P = IV.
Result of power calculation as 5.76 watts.
Discussion on the significance of current direction and its impact on the circuit.
Explanation of the current distribution through the resistors and its relation to their resistance values.
Conclusion summarizing the application of Ohm's law and Kirchhoff's laws to solve the circuit problem.
Transcripts
foreign
and we are doing another practice
problem and this one has to do with
resistors and current dividers and this
is in Irvine the 12th edition e 2.12
so here's our problem let's start with
reading the problem and looking at the
circuit so find currents i1 and I2 and
the power absorbed by the 40 kiloohm
resistor so let's just look at our
circuit here we have a current Source in
the middle and it's going around there's
two resistors one on each side and they
are in parallel so and we can see the
way that the current is labeled I want
is labeled going from bottom to top on
the left resistor and then the one on
the right is labeled from I2 is from top
to bottom on the right resistor
okay so now we've read the problem let's
write down what we need to find up here
so we're going to need to find I one
and we're gonna have to find I2 as well
and we'll also have to find power of the
40
um kilo Ohm resistor and we'll write
that down as well
so
we see that if current is Flowing here
current is going to flow in a loop so
it's going to go in this direction
and this is going to be our our source
of power and then we have two resistors
and we know resistors are always
absorbing power
so they can't Supply Power so it's one
thing this kind of makes sense
um the current is going to go in this
direction and just so that we understand
om La I'm going to put a positive and
negative here just to show the direction
because it's going through the resistor
in this direction the I2 if it were
going in the same direction it would be
positive to negative from the top but
actually because these two are in
parallel the voltages have to be the
same so if this is a positive voltage
relative to this top node it's going to
be the same polarity on the right and
that's
actually what's going to happen so
because the currents are set up
differently I'm actually going to define
something new I'm going to Define
um I'm going to call it
I be because I'm just redefining it and
we're going to call IB it's going to be
the negative of I2
okay and I'm going to do this because
when it's defined in the same direction
its polarity can be the same and I'm
going to go ahead and name this voltage
the voltage over here we're going to
call that VX
and it's going to be the same on both
sides
okay so now with this setup I'm going to
look at
one of our or apply crickets current law
and then we're going to write an
equation for it so let's apply
kirchhoff's current law here
we're going to look at this node and
look at the current coming into it 16
milliamps and then I've redefined it I'm
going to look at IB and I'm going to
write that here I be here and here is
going to be I one so if you write off
current law it's going to be 16
milliamps
has to be equal to
I 1 plus I B
all right
so now I need to find these current
values but I know the current is going
through a resistor whenever we see a
resistor we can think Ohm's law so let's
just write out Ohm's law so V equals i r
we know the resistance values those are
given
but we want to find the current so let's
just rewrite this in a different way so
we're going to write I equals
V over R so I want to know the currents
i1 and IB
and I know the resistances and actually
look I also know the voltages because I
named them they're both going to be the
same because these are in parallel so
the voltages over each of the resistors
is going to be VX so what we can do is
rewrite each of these so we can write i1
and right over here i1 equals VX over
40 kilo ohms
and I can write i b over here remember
we're working with IB
and I can write that as v x
over the resistance which is 120 kilo
ohms
all right so now I can plug those values
into each of these
equations or these variables and let's
rewrite that
so I'm going to get 16 milliamps
equals VX over 40
kilo ohms Plus
the X over 120
kilo ohms
all right we have one equation and one
variable VX that we need to solve for so
let's solve for that variable and we see
we want to put this in terms of 120 kilo
Ohms on the bottom so we can multiply
this by a 3 over 3 and then I'm going to
go ahead and move that over so let's see
we're going to get 16.
milliamps and we're going to multiply it
by 120 kilo ohms
and then here we're going to get a 3vx
plus a 1vx so it's going to be a 4 VX
and then I can divide through by
um
by four here and I'm just gonna do we're
gonna go old school I'm not even gonna
you know we're gonna divide this by four
and then 120 kilo ohms VX and so our VX
our final value
will be
480 volts
okay so this is like really a high
voltage
anyway it's just an example so this is
quite a high voltage
um but if you put that over here you'll
see that you'll get the the currents out
so actually let's let's calculate it
let's bring this back I'm going to
rewrite i1 over here
it's going to be VX
right over 40 kilo ohms but we just
found the X so we can do 480.
over 40
uh kilo ohms
so we're going to get uh 12 right and
then we're dividing by K so we're going
to Milli and then we know we're gonna be
in amps so 12 milliamps is our answer
and that's i1 we found that so we can
just write that up here 12 milliamps
next we can do uh IB IB
IB is VX over
120 kilo ohms
and so we're going to get 480 volts
over 120
ohms okay when we divide that out we are
going to get four
milliamps
so we know that IB is 4 milliamps and
that means that I 2 is going to be the
negative value of that so it's going to
be a negative 4 milliamps
so when you write it back here just make
sure to put the right notation so it's a
negative 4 milliamps and now we can go
and solve our last part which is the
power of the 40 kilo Ohm resistor so
just remember that P
equals IV
and so we have uh the voltage
is
480 volts and then the current is 12.
milliamps
if we do the math on that we're going to
get five seven six zero milliwatts why
don't we just write that as a
5.76 watts and then that is our power
through here
so these are all the things that we were
asked to find we found the currents for
i1 and I2 and the power through the 40
kilo Ohm resistor
do like what does this mean so it means
that the i1 which is over here was going
in the same direction and it had a
larger current value notice that these
resistor values this resistor on the
right is three times larger than the
other one but actually the current
through it was smaller in magnitude so
we had only four Milli ohms coming
through this one and 12 through here so
we can also look at the values and kind
of get a proportion of how much current
we could expect to go through each
and we also noticed that although I2 was
defined going downward it really makes
sense in terms of circuitry to have the
current go in the opposite direction so
when we have a problem like that we can
redefine our current direction I
redefined it as IB to go in the same
direction because if we have these in
parallel the current is going to go from
bottom to top in both of them so we
redefine that and then when we write our
answer out we just get a negative
current just means that the direction of
I not is against the arrow so it's the
opposite direction as the arrow is shown
in I2
then from there once we know the voltage
and the current then we can just
calculate the power and that's our
answer
so we've applied Ohm's law and for
perhaps current law to solve this
circuit and find both currents and the
power through one of the resistors
[Music]
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