Electric Current & Circuits Explained, Ohm's Law, Charge, Power, Physics Problems, Basic Electricity
Summary
TLDRThis 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
🔋 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.
🔌 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.
💡 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.
⚡ 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.
Mindmap
Keywords
💡Electric Current
💡Ohm's Law
💡Resistance
💡Voltage
💡Electron Flow
💡Electric Power
💡Coulombs
💡Amps
💡Milliamps
💡Watts
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
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