Introduction to Ohm's Law
Ohm's Law defines the fundamental relationship between voltage (V), current (I), and resistance (R) in an electrical circuit through the equation V = IR. Voltage is measured in volts, current in amps, and resistance in ohms. The law states:
- Voltage and current are directly proportional when resistance is constant.
- Current and resistance are inversely proportional when voltage is constant.
Calculating Current in Simple Circuits
Example: A 12-volt battery connected across a 4-ohm resistor.
- Using Ohm's Law: I = V / R = 12 V / 4 Ω = 3 amps.
Current flows conventionally from the battery's positive to negative terminal, opposite to electron flow.
Series Circuits and Total Resistance
When resistors are connected in series, total resistance (R_total) is the sum of individual resistances:
- R_total = R1 + R2 + R3
Example: Three resistors of 3 Ω, 4 Ω, and 5 Ω connected across a 60 V battery.
- R_total = 3 + 4 + 5 = 12 Ω
- Current: I = V / R_total = 60 V / 12 Ω = 5 amps (same through all resistors)
Voltage Drops Across Series Resistors
Calculate voltage drop across each resistor using V = IR:
- V1 = 5 A × 3 Ω = 15 V
- V2 = 5 A × 4 Ω = 20 V
- V3 = 5 A × 5 Ω = 25 V
Sum of voltage drops equals total battery voltage (15 + 20 + 25 = 60 V), demonstrating Kirchhoff's Voltage Law.
Kirchhoff's Voltage Law (KVL)
KVL states the sum of voltages around any closed loop is zero, balancing the energy supplied by the battery and consumed by circuit elements.
Parallel Circuits and Current Distribution
In parallel connections, voltage across each resistor is the same, but currents vary inversely with resistance.
Example: Three resistors of 3 Ω, 4 Ω, and 6 Ω connected in parallel across a 12 V battery.
- Current through R1: I1 = 12 V / 3 Ω = 4 A
- Current through R2: I2 = 12 V / 4 Ω = 3 A
- Current through R3: I3 = 12 V / 6 Ω = 2 A
Total current leaving battery: I_total = I1 + I2 + I3 = 9 A
Higher resistance means lower current, as reflected here.
For further details on calculating combined resistances in such configurations, see How to Solve Series-Parallel Resistor Circuits: Step-by-Step Guide.
Kirchhoff's Current Law (KCL)
KCL states that the total current entering a junction equals the total current leaving it.
Example: 9 A current enters a junction; if 4 A leaves one branch, 5 A must leave the other to conserve current.
Summary
- Ohm's Law is fundamental for calculating voltage, current, and resistance.
- Series circuits have the same current and additive resistances.
- Parallel circuits share the same voltage, but currents divide based on resistance.
- Kirchhoff's Laws assist in analyzing complex circuits ensuring energy and charge conservation.
This knowledge equips you to solve practical circuit problems and understand electrical energy flow effectively.
For a deeper foundational understanding, consider reviewing Understanding Electricity: The Basics of Current, Potential Difference, and Resistance.
in this video we're gonna talk about Ohm's law so what is Ohm's law Ohm's law describes the relationship between
voltage current and resistance perhaps you've seen this equation V is equal to I times R V stands for voltage I stands
for the current r is resistance voltage is measured in the units of volts I which represents the current is measured
in amps and are the resistance is measured in ohms now you need to know that as the voltage in a circuit
increases the current will increase provided that the resistance stays the same if the resistance goes up the
current will go down if the voltage is held constant so voltage and current they are proportional to each other and
resistance and current they're inversely related to each other now let's work on a practice problem
let's say if we have a 12 volt battery connected across a four ohm resistor what is the current flowing in this
circuit conventional current flows from the positive terminal the battery to the negative terminal of the battery this is
the opposite direction to electron flow so to find the current in a circuit we can use Ohm's law V is equal to IR
so the voltage is 12 we're looking for the current the resistance is 4 so we need to solve for the variable I let's
divide both sides by 4 12 divided by 4 is 3 so the current is going to be 3 amps
now let's say if we have three resistors connected in series like this let's say this is r1 r2 and r3 and it's connected
across a 60 volt battery now let's say that r1 has a value of 3 ohms and r2 is 4 ohms and r3 is 5 ohms so what is the
current flowing in a circuit in order to find the current flowing in a circuit where the resistors are connected in
series you need to find the total resistance and the total resistance is going to be r1 plus r2 plus r3 you just
need to add the values of the three resistors so 3 plus 4 plus 5 that's going to give us 12 so the total
resistance in a circuit is 12 ohms next you need to calculate the SIRT on the current so we could use the formula V is
equal to IR so V is 60 that's the voltage across the three resistors we're looking for the current and then the
total resistance is 12 so you can treat this as if it's one big resistor and you have a 60 volt battery source across a
12 ohm resistor what is the current in that resistor so what we need to do is divide both sides by 12 to get the
current by itself 60 divided by 12 is 5 so we have a current of 5 amps flowing in this circuit now once we have the
current we can calculate the voltage drop across each resistor what is the voltage drop
across the first resistor now in this series circuit the current that flows in a circuit is the same as the current
flowing through r3 r2 and r1 because there's only one path for the current to flow it's going to be the same five amps
so to find the voltage across the first resistor we can use the current that flows through the first resistor times
the resistance to the value of that resistor so we're using Ohm's law but in a different way I 1 is going to be the
same as I because that 5 amp current is flowing through each resistor but r1 is different r1 is going to be 3 so it's 5
times 3 so we have 15 volts across r1 now what about across r2 what is the voltage across r2 well we could follow
the same pattern so we can say v2 is equal to i2 times r2 so the current is still going to be 5 amps but this time
the resistance is 4 ohms so 5 times 4 that's going to give us 20 so we have 20 volts across r2 now across R 3 it's
going to be v3 is equal to I 3 times r3 so I 3 is going to be the same as i2 and i1 so that's 5 amps r3 is 5 so 5 times 5
is 25 so notice that if you add up 15 plus 20 plus 25 it gives you 60 and so the voltage of the battery is equal to
the sum of all of the voltage drops across those resistors and there's something called Kirchhoff's voltage law
which basically states that as you go around a circuit in a loop the total voltage will
be zero and make sense because the battery it increases the energy of the circuit because it supplies energy to
the circuit so increases it by 16 the resistors consume energy from the circuit so they decrease it does they
have a negative value so if you add a positive 60 with a negative 15 negative 20 and negative 25 you get zero because
the energy that flows into a circuit must equal the energy that comes out of your circuit thus kirchoff's voltage law
it always applies whenever you have a closed loop so the sum of all the voltages in a closed loop will always
add up to zero now what's going to happen if we connect three resistors in a parallel circuit let's calculate the
current in such a circuit in the series circuit the current flowing through the resistors that are
connected in series is the same because the current only has one path and what you could flow in the parallel circuit
the current has multiple paths and so it could vary however notice that whenever resistors are connected in parallel the
voltage across those resistors is the same so let's say if we have in this case a 12 volt battery each resistor is
connected across that 12 volt battery and so all of them have 12 volts across their terminals let's call this r1 r2
and r3 and so let's say that r1 has a value of 3 ohms and r2 is going to be 4 ohms and r3 is going to be 6 ohms what
is the current flowing through each resistor so we can use this formula V 1 is equal to i1 times R 1 so remember in
a parallel circuit the voltage across the resistors connected in parallel is the same but in the series circuit the
current flowing and resistors that are connected in series will be the same in this case V 1 is 12 because we have 12
volts connected across R 1 and to find the current flowing through this resistor we need to use that formula R 1
s 3 so the current is going to be 12 divided by 3 so we have a current of 4 amps flowing through R 1 now let's do
the same for R 2 so let's use the formula V 2 is equal to i2 times R 2 it's basically owns a labo of different
subscripts so V 2 is still 12 and R 2 is not 4 so it's going to be 12 divided by 4 which will give us a current of 3 amps
now to calculate I 3 it's going to be 12 the by 6:00 following the same pattern and
so that's a current of two amps now notice that as the resistance increases the current decreases as we mentioned in
the beginning of this video here notice that our one has the lowest value and it has the highest current our three has
the highest value but it has a lowest current so as you increase the resistance the current decreases if we
increase it to six the current decreases the two and if we decrease the resistance the current will increase if
we decrease it to three the current goes up to four provided that the voltage is held constant and so you'll see this
relationship in a parallel circuit as you can see it here or in a series circuit - you can see that relationship
there as well now what is the current that leaves the battery how can we determine the current that's leaving the
battery the total current in the circuit which we'll call I T that leaves the battery it's going to be the sum of the
individual currents so it's going to be four plus three plus two four plus three plus two is nine so the total current
here is nine amps now let's focus on this point what is the current that is flowing through that branch right here
along let's say this wire what is the current in that region so now we need to use something called Kirchhoff's to
current law we saw in a last example that kirchoff's voltage law which basically states that the sum of all the
voltages around the loop adds up to zero while kirchoff's current law is very similar the current that enters the
junction is equal to the current that leaves the junction now let's draw a picture
so we have 9 amps of current flowing to this point and we have 4 amps that's leaving it now the current that is
flowing to a junction must equal the total current that is leaving the junction so we have to have current
leavin in this direction and it has to be 5 amps because 5 plus 4 is 9 so we have a total of nine amps of current
that answers disjunction and 9 amps a current that leaves it and so that's the basic idea behind Kirchhoff's current
law so we have five amps flow into the right in this direction now out of those five amps three amps is going this way
so that means the other two amps flows in this direction and you can see how it's like a river splitting off into
three directions so in this section we still have two amps of the current that's traveling here and then when it
joins up with a three amp current three amps you will add up to five and so we have five amps flowing in this region
and then the five and four will get together and so we're gonna have a total of nine amps of current flowing in this
region and so as you can see the current that's going this way it's gonna be the same as this current here not amps and
so hopefully this all makes sense it's I want to give you a basic idea of how to use Ohm's law in a simple circuit in a
series circuit and also in a parallel circuit so that's it for this video hopefully you found it to be helpful and
they gave you a good understanding of kirchoff's voltage law and its current law as well thanks for watching
Ohm's Law states that voltage (V) is equal to the current (I) multiplied by the resistance (R), expressed as V = IR. To calculate current, rearrange the formula to I = V / R. For example, if a 12-volt battery is connected to a 4-ohm resistor, current I = 12 V / 4 Ω = 3 amps.
In a series circuit, total resistance is the sum of all individual resistors (R_total = R1 + R2 + ...). The current flowing through each resistor is the same. For instance, with resistors of 3 Ω, 4 Ω, and 5 Ω in series connected to a 60 V battery, total resistance is 12 Ω and current I = 60 V / 12 Ω = 5 amps through each resistor.
Voltage drop across each resistor in series is found using V = IR, where I is the current through the resistor. Using the previous example with 5 A current: V1 = 5 A×3 Ω=15 V, V2=5 A×4 Ω=20 V, and V3=5 A×5 Ω=25 V. The sum of these voltage drops equals the total supply voltage, demonstrating Kirchhoff's Voltage Law.
Kirchhoff's Voltage Law states that the sum of all voltages around a closed loop equals zero. It ensures that the energy supplied by sources in a circuit equals the energy consumed by circuit elements, allowing you to verify voltage balances and solve complex circuits accurately.
In parallel circuits, voltage across each resistor is the same, but current divides inversely with resistance. For example, three resistors of 3 Ω, 4 Ω, and 6 Ω across a 12 V battery carry currents of 4 A, 3 A, and 2 A respectively. Higher resistance yields lower current, and total current is sum of individual currents.
Kirchhoff's Current Law states that the total current entering a junction equals the total current leaving it, ensuring conservation of charge. For example, if 9 A enter a junction, and 4 A leaves one branch, then 5 A must leave the other branch to maintain current balance.
These laws provide foundational relationships to calculate voltage, current, and resistance in circuits, and analyze complex arrangements by ensuring conservation of energy and charge. Using them, you can determine values across components, verify circuit function, and design efficient circuits with predictable behavior.
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