Practice Problem: Solving a Circuit Using Ohm's Law and Kirchhoff's Current Law

Summary

This 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

00:00

🔌 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.

05:04

🔍 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.

10:07

📚 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.

Keywords

💡Resistors

Resistors are components in electrical circuits that oppose the flow of current. They are essential for controlling the amount of current in a circuit and are measured in ohms. In the video, resistors are part of a parallel circuit and are used to demonstrate the current divider principle, where the total current is divided between the two resistors based on their resistance values.

💡Current Dividers

A current divider is a principle in electrical circuits where a single current source is split into multiple paths, each with its own resistor. The amount of current flowing through each path is determined by the resistance of the resistors involved. In the video, the concept of current dividers is central to solving the problem presented, which involves calculating the currents through two resistors in parallel.

💡Kirchhoff's Current Law

Kirchhoff's Current Law, also known as Kirchhoff's First Law, states that the total current entering a junction in a circuit is equal to the total current leaving the junction. In the video, this law is applied to a node in the circuit to establish an equation that helps solve for the unknown currents in the circuit.

💡Ohm's Law

Ohm's Law is a fundamental principle in electrical engineering that relates voltage (V), current (I), and resistance (R) in a circuit with the formula V = I * R. It is used to calculate the current flowing through a resistor when the voltage across it is known, or vice versa. In the script, Ohm's Law is applied to find the currents through the resistors once the voltage across them is determined.

💡Parallel Circuit

A parallel circuit is a type of electrical circuit where multiple components are connected in parallel, meaning they have the same voltage across them but may have different currents flowing through each component. In the video, the problem involves a parallel circuit with two resistors, and understanding this configuration is key to solving the problem.

💡Voltage (VX)

Voltage, denoted as VX in the script, is the electrical potential difference between two points in a circuit. It is measured in volts and is crucial in determining the direction and magnitude of the current flow. In the context of the video, VX represents the voltage across each resistor in the parallel circuit, which is the same due to the nature of parallel connections.

💡Current (I1, I2, IB)

Current is the flow of electric charge in a circuit, measured in amperes (amps). In the video, the task is to find the currents I1 and I2 flowing through two resistors in a parallel circuit. Additionally, IB is introduced as a redefined current to simplify the analysis, representing the negative of I2 for consistency in direction.

💡Power Absorption

Power absorption refers to the rate at which electrical energy is consumed or transformed in a circuit component, such as a resistor. It is calculated using the formula P = IV, where I is the current and V is the voltage across the component. In the video, the power absorbed by the 40 kiloohm resistor is one of the quantities to be determined, indicating the amount of energy being dissipated as heat.

💡Kilohms

Kilohms is a unit of measurement for electrical resistance, equal to one thousand ohms. In the script, the resistance values of the resistors are given in kilohms, specifically 40 kiloohms and 120 kiloohms, which are used in the calculations to determine the currents and voltages in the circuit.

💡Milliamps

Milliamps is a unit of electric current, equal to one thousandth of an ampere. In the video, the currents I1 and IB are calculated in milliamps, which is a common unit for expressing small currents in electrical circuits. The script uses this unit to express the results of the current calculations.

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]