Electric Current & Circuits Explained, Ohm's Law, Charge, Power, Physics Problems, Basic Electricity

Summary

This educational video script covers fundamental electrical concepts, including the nature of electric current, Ohm's Law, and the calculation of charge, current, voltage, resistance, and power. It explains the direction of conventional current versus electron flow, defines current as the rate of charge flow, and introduces Ohm's Law as the relationship between voltage, current, and resistance. The script also discusses the formula for electric power and its different forms, and includes practical problem-solving examples to illustrate these concepts.

Takeaways

• 🔋 The conventional current flows from the positive to the negative terminal of a battery, analogous to water flowing from a high to a low position.
• ⚡ Current is defined as the rate of charge flow, mathematically expressed as charge (q) divided by time (t), with the unit of current being the ampere (amp).
• ⚛️ Electrons actually flow from the negative terminal to the positive terminal, which is the opposite direction of conventional current.
• 🔗 Ohm's Law (V=IR) describes the relationship between voltage (V), current (I), and resistance (R), where voltage is the product of current and resistance.
• 📈 Increasing voltage while keeping resistance constant increases current, and increasing resistance decreases current, showing direct and inverse relationships respectively.
• 🚦 The concept of resistance in a circuit can be compared to the number of lanes on a highway, where more lanes allow for easier flow (less resistance).
• 💡 Electric power is calculated as the product of voltage and current (P=VI), and can also be expressed as I^2R or V^2/R, with power measured in watts.
• ⏱️ To calculate the charge that passes through a circuit, multiply the current by the time in seconds, ensuring unit consistency.
• 💸 The cost to operate an electrical device can be determined by calculating the energy consumption over a period and multiplying by the cost per kilowatt-hour.
• 🔌 The internal resistance of a device can be found using Ohm's Law by rearranging the formula to solve for resistance (R=V/I).

Q & A

What is the direction of conventional current flow?

Conventional current flows from the positive terminal to the negative terminal, from high voltage to low voltage.

How is electric current defined?

Electric current is defined as the rate of charge flow, which is the charge divided by time, expressed as I = Q/t or delta Q over delta t.

What is the unit of electric charge and what is the charge of a single electron?

The unit of electric charge is the coulomb. A single electron has a charge of 1.6 times 10 to the negative 19 coulombs.

What is Ohm's Law and how is it expressed mathematically?

Ohm's Law describes the relationship between voltage, current, and resistance. It is expressed as V = IR, where V is voltage, I is current, and R is resistance.

How do voltage and current relate when resistance is constant?

When resistance is constant, an increase in voltage will increase the current, and an increase in resistance will decrease the current.

What is electric power and how is it calculated?

Electric power is the rate at which energy is transferred, calculated as the product of voltage and current (P = VI), or I^2R, or V^2/R.

How can you calculate the charge that passes through a circuit given the current and time?

The charge that passes through a circuit can be calculated by multiplying the current (in amperes) by the time (in seconds), Q = I * t.

How many electrons are represented by 2736 coulombs of charge?

2736 coulombs of charge represents approximately 1.71 times 10 to the 22 electrons, considering one electron has a charge of 1.6 times 10 to the negative 19 coulombs.

If a 9-volt battery is connected across a 250-ohm resistor, what is the current passing through the resistor?

Using Ohm's Law, the current passing through the resistor is calculated as I = V/R, which is 9 volts / 250 ohms, resulting in 0.036 amps or 36 milliamps.

What is the cost to operate a 1.8-watt light bulb for a month if electricity costs 11 cents per kilowatt-hour?

The cost to operate a 1.8-watt light bulb for a month is approximately 14 cents, calculated by converting watts to kilowatts, multiplying by the number of hours in a month, and then by the cost per kilowatt-hour.

Outlines

00:00

🔋 Basic Concepts of Electric Current and Ohm's Law

This paragraph introduces the fundamental concepts of electric current and Ohm's Law. It explains the direction of conventional current, which is from the positive to the negative terminal of a battery, and contrasts it with the actual flow of electrons from the negative to the positive terminal. The paragraph defines electric current as the rate of charge flow, measured in amperes (amps), and explains that 1 amp is equivalent to 1 coulomb per second. It also discusses the charge of an electron, which is 1.6 x 10^-19 coulombs. Ohm's Law is introduced as the relationship between voltage, current, and resistance, with the formula V = IR, where V is voltage, I is current, and R is resistance. The effects of changing voltage and resistance on current are discussed, highlighting the direct relationship between voltage and current, and the inverse relationship between resistance and current.

05:00

🔌 Calculations Involving Current, Charge, and Power

This paragraph focuses on practical calculations related to electric current, charge, and power. It begins with a problem involving a current of 3.8 amps flowing through a wire for 12 minutes, asking how much charge passes through any point in the circuit. The calculation involves converting time from minutes to seconds and then multiplying the current by time to find the charge in coulombs. The paragraph then extends the discussion to determine the number of electrons represented by this charge, using the charge of a single electron. It also covers a problem involving a 9-volt battery connected to a 250-ohm resistor, calculating the current through the resistor using Ohm's Law and then calculating the power dissipated by the resistor using the formula P = I^2R. The paragraph concludes with a discussion on the power delivered by the battery, ensuring that the power delivered equals the power dissipated in the circuit.

10:01

💡 Resistance and Power Calculations for a Light Bulb

The third paragraph delves into the calculations related to a 12-volt battery connected to a light bulb drawing 150 milliamps of current. It begins by determining the electrical resistance of the light bulb using Ohm's Law, converting milliamps to amps and then calculating the resistance in ohms. The paragraph then calculates the power consumed by the light bulb using two different formulas: P = VI and P = I^2R, both yielding the same result. It also addresses the cost of operating the light bulb for a month, given the cost of electricity and the power consumption of the bulb. The calculation involves converting the power from watts to kilowatts and then determining the energy consumption over a month, followed by the cost based on the electricity rate.

15:02

⚡ Voltage, Resistance, and Power in an Electric Motor

This paragraph discusses the calculations for an electric motor that uses 50 watts of power and draws a current of 400 milliamps. It starts by determining the voltage across the motor using the power formula P = VI, converting milliamps to amps and then calculating the voltage. The internal resistance of the motor is then calculated using Ohm's Law, V = IR. The paragraph concludes with the calculation of the power consumed by the motor, which is confirmed using both the power formula and the product of current and resistance.

🕰 Time-Based Current and Power Calculations

The final paragraph deals with time-based calculations of electric current and power. It presents a scenario where 12.5 coulombs of charge flow through a 5-kilo ohm resistor in eight minutes. The paragraph explains how to calculate the electric current by dividing the charge by the time converted to seconds. The resulting current is then used to calculate the power consumed by the resistor using the formula P = I^2R. The voltage across the resistor is also determined using Ohm's Law, V = IR. The calculations are detailed, providing a clear understanding of how to work with time-dependent electrical quantities.

Keywords

💡Electric Current

Electric current refers to the flow of electric charge, typically carried by electrons, through a conductor such as a wire. In the video, current is defined as the rate of charge flow, mathematically expressed as the charge divided by time (I = Q/t). It is measured in amperes (amps), where 1 amp is equal to 1 coulomb per second. The script uses the analogy of water flowing from a high position to a low position to explain the direction of conventional current, which flows from the positive to the negative terminal of a battery.

💡Ohm's Law

Ohm's Law is a fundamental principle in electricity that establishes the relationship between voltage (V), current (I), and resistance (R) in a circuit. It is expressed as V = IR, indicating that the voltage across a conductor is directly proportional to the current flowing through it and inversely proportional to the resistance. The video uses Ohm's Law to solve various problems, such as calculating the current through a resistor given a voltage and resistance, and to explain how increasing voltage increases current and how increasing resistance decreases current.

💡Resistance

Resistance is a property of materials that opposes the flow of electric current. It is measured in ohms and is influenced by the material's composition, length, and cross-sectional area. In the video, resistance is used to explain how it affects the current flow in a circuit. The script provides an analogy comparing a one-lane highway to a seven-lane highway to illustrate how resistance impacts the flow of electric current, similar to how traffic flow is affected by the number of lanes available.

💡Voltage

Voltage, also known as electric potential difference, is the force that pushes electric charge through a circuit. It is measured in volts and is the cause of electric current. The video script mentions voltage in the context of Ohm's Law and when calculating the power delivered by a battery. For example, when a 9-volt battery is connected across a resistor, the voltage is used to determine the current flowing through the resistor.

💡Electron Flow

Electron flow is the actual movement of electrons through a conductor, which is opposite to the conventional current flow. While conventional current is assumed to flow from positive to negative, electrons, which carry a negative charge, actually flow from negative to positive. The video script clarifies this distinction and uses it to explain the direction of current in a circuit, emphasizing that electrons originate from the negative terminal and move towards the positive terminal.

💡Electric Power

Electric power is the rate at which electrical energy is transferred by an electric circuit. It is measured in watts and can be calculated using the formula P = VI, where P is power, V is voltage, and I is current. The video script discusses electric power in the context of calculating the power dissipated by a resistor and the power delivered by a battery. It also introduces alternative forms of the power equation, such as P = I^2R and P = V^2/R, which are used to solve problems involving power in circuits.

💡Coulombs

A coulomb is the unit of electric charge named after Charles-Augustin de Coulomb, the French physicist. In the video, coulombs are used to measure the quantity of electric charge, with the script mentioning that the charge of an electron is 1.6 x 10^-19 coulombs. The video uses the concept of coulombs to calculate the total charge passing through a point in a circuit over a given time, which is then used to determine the number of electrons represented by that charge.

💡Amps

Ampere, or amp, is the unit of electric current. The video script defines 1 amp as the flow of 1 coulomb of charge per second. Amps are used throughout the video to calculate and discuss the flow of electric charge in various circuits, such as when determining the current that flows through a resistor connected to a battery or when calculating the current from a given charge and time.

💡Milliamps

Milliampere, or milliamp, is one-thousandth of an ampere. The video script converts current measurements from milliamps to amps for calculations, such as when determining the current flowing through a light bulb connected to a battery. This conversion is necessary because milliamps provide a more precise measurement for small currents, and the script uses this unit to demonstrate how to convert between different units of current.

💡Watts

Watt is the unit of electric power, named after James Watt, the inventor and mechanical engineer. The video script uses watts to measure the power delivered by a battery and the power dissipated by a resistor. It explains that one watt is equal to one joule per second, emphasizing the rate at which energy is transferred. The script calculates power in various scenarios, such as when a resistor is connected to a battery, to demonstrate the application of the power equation.

Highlights

Conventional current flows from positive to negative terminal, opposite to electron flow.

Current is the rate of charge flow, defined as charge divided by time (delta Q/delta t).

The unit of current is the ampere (amp), where 1 amp = 1 coulomb per second.

Electrons have a charge of 1.6 x 10^-19 coulombs each.

Ohm's Law (V = IR) describes the relationship between voltage, current, and resistance.

Increasing voltage with constant resistance increases current; increasing resistance decreases current.

Electric power is the product of voltage and current, with three forms: P = VI, P = I^2R, P = V^2/R.

Power is measured in watts, where 1 watt = 1 joule per second.

Calculating electric charge involves multiplying current by time in seconds.

The number of electrons is proportional to the amount of charge.

Using Ohm's Law to calculate current when voltage and resistance are known.

Electric power dissipated by a resistor can be calculated using P = I^2R.

The power delivered by a battery equals the power absorbed by the resistor in a simple circuit.

Calculating electrical resistance involves rearranging Ohm's Law to R = V/I.

Electrical power consumption can be calculated using P = VI or P = I^2R.

Operating cost of an electrical device is determined by its power consumption and the cost of electricity.

Electric current can be calculated from charge and time using the formula I = Q/t.

Power consumed by a resistor can be found using P = I^2R, and voltage across it using V = IR.

Transcripts

in this video we're going to go over a

few basic equations and work on some

practice problems involving electric

current and ohm's law

so let's say if we have a battery

the long side of the battery is the

positive terminal and here's a resistor

conventional current

states that

current flows from the positive terminal

to the negative terminal current flows

from high voltage

to low voltage

that's conventional current the same way

as water flows from a high position to a

low position

electron flow is the opposite

in reality we know that electrons they

emanate from the negative terminal

and flow towards the positive terminal

so just keep that in mind

but now let's talk about current

conventional current which is the flow

of positive charge

current

is defined as

it's basically the rate of charge flow

it's charge divided by time

or delta q over delta t

q is the electric charge measured in

coulombs

and t is the time in seconds

the unit 4 current is the amp

so 1 amp

is equal to 1

per second

and electric charge

is associated with the quantity of

charged particles

an electron

has a charge that's equal to 1.6 times

10

to the negative 19 coulombs

and it's negative

now there are some other equations that

we need to talk about

and one of them is ohm's law

which describes the relationship between

voltage current and resistance

v is equal to ir voltage is the product

of the current and resistance the

resistance

is measured in ohms that's the unit of

resistance

now

keeping the resistance constant

if you were to increase the current what

do you think the effect will be on the

voltage

or rather if you increase the voltage

what is the effect on the current

increasing the voltage will increase the

current

and what about increasing the resistance

what effect will that have on the

current

if you increase the resistance the

current will decrease but if you were to

increase the voltage the current will

increase

the voltage and the current are directly

related the resistance and the current

are inversely related

the more resistance you have in a

circuit it's harder for current to flow

it's just

it's not going to flow as well think of

a high wing

if you have a seven-lane highway it's

going to be easy for cars to flow

through it

as opposed to a one-lane highway a

one-lane highway has more resistance so

less cars can flow through it the cars

being the flow of electric current

but if you decrease resistance if you

add more lanes to the road

more cars could flow so there's more

current

the next equation you need to be

familiar with is electric power

electric power is the product of the

voltage and the current

now this three forms to this equation

so if you replace v with ir

you can get the second form

which is i squared times r

and if you replace i with v over r you

can get the third form which is

v squared over r

so power is equal to voltage times

current or i squared times r or v

squared over r

power is measured in watts

power is the rate at which energy can be

transferred

one watt

is equal to one joule per second

so these are some things to know

now let's work on some problems

a current of 3.8 amps

flows in the wire for 12 minutes

how much charge passes through any point

in the circuit during this time

so we have the current

it's 3.8 amps

and we have the time which is 12 minutes

how can we calculate the electric charge

well we know that high

is q divided by t

so to solve for the electric charge q

it's i times t

so we have to multiply but we need to be

careful with the units though

t

is the time in seconds

so let's convert 12 minutes into seconds

each minute

equates to 60 seconds

so we've got to multiply by 60.

notice that the unit minutes cancel

12 times 60 is 720

so t is 720 seconds

now let's calculate q

so it's equal to i the current which is

3.8 amps

multiplied by

720 seconds

so the electric charge

is

2736 columns

now what about part b

how many electrons would this represent

so if you have the charge you can easily

convert it to number of electrons

let's start with this number

now it turns out that one electron

has a charge of 1.6 times

10 to the negative 19 coulombs i'm gonna

have to worry about the negative sign

so if we divide these two numbers

27 36 divided by 1.6 times 10 to the

negative 19

this will give you the number of

electrons

and so that's going to be about 1.71

times 10 to the 22

electrons

so keep in mind

the amount of charge is proportional to

the number of electrons

so that's it for this problem

number two a nine volt battery is

connected across a 250 ohm resistor

how much current passes through the

resistor

well

we can begin by drawing a circuit here's

the battery

and here is the resistor

so we have

a 9 volt battery

and a 250 ohm resistor

what equation do we need to calculate

the electric current

the equation that we can use is ohm's

law

v is equal to ir the voltage is 9 and

the resistance is 250 so solving for i

let's divide both sides by 250.

so the current is equal to the voltage

divided by the resistance

so

9 volts divided by 250 ohms

is equal to

0.036 amps

now if you want to convert amps into

milliamps multiply by a thousand

or move the decimal three units to the

right so this is equivalent to 36

milliamps

part b

how much power is dissipated by the

resistor

so what equation can we use here

well let's use this equation power is

equal to i squared times r

the current that flows through the

resistor

is .036 amps

and we need to square it and the

resistance is 250 ohms

0.036 squared is

0.001296 and if we multiply that by 250

this is going to give us 0.324

watts

which is equivalent to 324 milliwatts

part c

how much power is delivered by the

battery

well let's use this equation p is equal

to v times i

the voltage of the battery is 9 volts

and the current that the battery

delivers

is the same as the current that flows

through the resistor which is

0.036 amps

9 times 0.36

is equal to the same thing point

watts and it makes sense

everything has to be balanced

the amount of power delivered by the

battery should be equal to the amount of

power dissipated or absorbed by the

resistor

because that's the there's only two

elements in the circuit the battery

delivers energy the resistor absorbs it

so if they're the only two circuit

elements

the amount of power transferred has to

be equal

number three

a 12 volt battery is connected to a

light bulb and draws 150 milliamps of

current

what is the electrical resistance of the

light bulb

let's draw a circuit

so let's say

this is the light bulb

and we have a battery

connected to it

and that's the positive terminal here's

the negative terminal

and electric current flows from the

positive side

to the negative side but electrons will

flow in the opposite direction

now let's make a list of what we know

so the voltage is 12.

the current is 150 milliamps but we'll

need that in amps

so we can divide that by a thousand or

move the decimal three units to left

so that's equivalent to point 15

amps

so now we could find the electrical

resistance using ohm's law

v is equal to ir

so the voltage is 12 the current is 0.15

and let's solve for r

we can do that by dividing both sides by

0.15

so the electrical resistance is equal to

the voltage

divided by the current

12 divided by 0.15

is 80.

so the internal resistance

of the light bulb is 80 ohms

now how much power does it consume

well we can use p is equal to vi

the voltage across the

light bulb is equal to the voltage of

the battery

that's 12

and the current delivered by the battery

is equal to the current absorbed by the

resistor

so that's 0.15

so 12 times 0.15 that's uh

1.8

watts

now we can also use i squared times r so

we can take the current which is point

fifteen

square it and then multiply by the

resistance which is eighty

point fifteen squared times eighty

will give us the same answer 1.8 watts

so you can use both techniques to

calculate the electrical power

now what about part c

how much will it cost to operate this

bulb for a month

if the cost of electricity

is 11 cents per kilowatt hour

well

we know the power that it uses is 80

watts

let's find out how much energy it uses

in a month then we can find out the cost

now i do have to make a small correction

the power is 1.8 watts and now 80 watts

so let's go ahead and begin with that

so first we need to convert watts into

kilowatts

we need to find the energy in kilowatt

hours

energy is basically power multiplied by

time power is energy over time

electric power is the rate at which

energy is transferred

now to convert watts to kilowatts let's

divide by a thousand

there's a thousand watts per kilowatt

now we need to multiply by the number of

hours

the total time that the light bulb is

going to be operating

is for one month

and there's 30 days

in a month on average

and there's about

24 hours per day

so notice that the unit months cancel

and the unit days cancel as well

leaving us with kilowatt times hours

so this will give us

the amount of energy being consumed in

one of to find the cost

let's multiply by

11 cents per kilowatt hour

and so now

the unit kilowatts will cancel

and the unit hours will cancel as well

so now all we need to do is just the

math

so it's 1.8 divided by a thousand

times 30 times 24

times 0.11

so it's only going to cost

14 cents

to operate

the light bulb for a month

number four a motor uses 50 watts of

power and draws a current of 400

milliamps what is the voltage across the

motor

so we have the power which is 50 watts

it's always good to make a list of what

you have

and the current is 400 milliamps we want

that in amps so we got to divide it by a

thousand which is 0.4 amps

so what is the voltage

well

electrical power is equal to voltage

times current

so p is 50 we're looking for v

and i is 0.4

so we need to divide both sides by 0.4

so 50 divided by 0.4

is 125.

so that's the voltage it's 125 volts

now what about part b what is the

internal resistance of the motor

well let's use ohm's law

v is equal to i r

so v is 125

i is 0.4

and let's find r so let's divide both

sides by 0.4

125 divided by 0.4

is equal to 312.5

ohms

so as you can see

these two equations are very important p

is equal to vi and v equals ir they're

very useful in solving common problems

number five

twelve point five columns of charge

flows through a five kilo ohm resistor

in eight minutes

what is the electric current that flows

to the resistor

so we have the charge q

it's 12.5

coulombs

and we have the electrical resistance

which is uh 5 kilo ohms

and we have the time eight minutes

how can we use this information to

calculate the electric current

well the electric current is the ratio

or really it's the change in the

electric charge

divided by the change in time

so it's the rate of charge flow it's how

much

charge flows per second

which means that we need to convert

eight minutes into seconds

so we got to multiply it by 60 seconds

6 times 8 is 48 so

60 times 8 is 480 you just gotta add the

zero so now we can find the electric

current

the charge that flows through any given

point is 12.5 coulombs

and let's divide that by 480 seconds

keep in mind one amp is one clone per

second

12.5 divided by 480

is .026

coulombs per second

or simply 0.026 amps

which is equivalent to 26 milliamps

now let's calculate how much power is

consumed

by the resistor so let's uh

make some space before we do that

what equation would you use

now we don't have the voltage so

let's use an equation that contains only

current and resistance

the current is 0.026 amps

and let's square that number and the

resistance was 5 kilo ohms

a kilo ohm is a thousand ohms so five

kilo ohms is five thousand ohms

point zero two six squared

is very small it's like six point seven

six times ten to the minus four

and if we multiply that by 5000

this is going to equal 3.38

watts

so that's how much power is consumed by

this resistor

if we want to find the voltage we can

it's simply equal to

i times r

it's .026 amps

times the resistance of 5000.

that's about 130 volts so that's the

voltage across the resistor and this is

the electrical power

which is what we want