V R And I In Parallel Circuits Answer Key? Top Answer Update

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To end the discussion of series-parallel circuits, I would like to post this last remaining topic, which is about Ohm’s law of series-parallel circuits for currents and voltages. I didn’t even mention in my previous topics how to deal with its currents and voltages in relation to this type of circuit connection.

Ohm’s law in series-parallel circuits

Ohm’s law in series-parallel circuits – current

The total current of the series-parallel circuits depends on the total resistance that the circuit presents when switched across the voltage source. Current flows throughout the circuit and splits to flow through parallel branches. In the case of a parallel branch, the current is inversely proportional to the resistance of the branch – that is, the greater current flows through the least resistance and vice versa. Then the current sums up again after flowing in another circuit branch identical to the current source or total current.

The total circuit current is the same at each end of a series-parallel circuit and is equal to the current flowing through the voltage source.

Ohm’s law in series-parallel circuits – voltage

Also, the voltage drops across series-parallel circuits occur in the same way as in series and parallel circuits. In series parts of the circuit, the voltage drop depends on the individual values ​​of the resistors. In parallel parts of the circuit, the voltage on each leg is the same and carries a current that depends on the individual values ​​of the resistors.

In the case of the following circuit, the voltage across the series resistor forming one branch of the parallel circuit divides the voltage across the parallel circuit. If in the case of the single resistor in a parallel branch the voltage across it is the same as the sum of the voltages of the series resistors.

The sum of the voltages across R3 and R4 is the same

than the voltage across R2.

Finally, the sum of the voltage drop across each path between the two terminals of the series-parallel circuit is equal to the total voltage applied to the circuit.

Let’s have a very simple example of this calculation for this topic. Considering the following circuit with its given values, let’s calculate the total current, current and voltage drop across each resistor.

What is the total current, current and voltage across each resistor?

 Here is the simple calculation of the above circuit:

a. First calculate the total resistance of the circuit:

The equivalent resistance for R2 and R3 is:

R2-3 = 25X50/ 25+50 = 16.67 ohms

R total = 30 ohms + 16.67 ohms = 46.67 ohms

b. Calculate the total current using Ohm’s law:

I1 = 120V / 46.67 ohms = 2.57 amps. Since R1 is in series, the total current for that path is the same.

c. Calculating the voltage drop for R1:

VR1 = 2.57A x 30 ohms = 77.1 volts

i.e. Calculate the voltage drop across R2 and R3.

Since the equivalent resistance for R2 and R3 is 16.67 ohms as calculated above, we can now calculate the voltage across each branch.

VR2 = VR3 = 2.57 amps x 16.67 ohms = 42.84 volts

e. Finally, we can now calculate the individual current for R2 and R3:

I2 = VR2 / R2 = 42.84 volts / 25 ohms = 1.71 amps.

I3 = VR3 / R3 = 42.84 volts / 50 ohms = 0.86 amps.

You can also check that the currents in each path of the parallel branch are correct by adding their currents:

I1 = I2 + I3 = 1.71A + 0.86A = 2.57A. This is the same as calculated above. Therefore we can say that our answer is correct.

Bottom up!

A short circuit occurs when there is an undesirably low resistance in or around a particular circuit.

A short circuit can occur when a piece of metal falls across a resistor. Remember that current takes the path of least resistance. So the current does not flow through the resistor, but through the metal. If the resistance is lower, the current will be higher and generate more heat in the circuit.

An unwanted short in a circuit can cause damage to other components and cause the resistor to burn out or the wire to burn out, among other things.

Circuit parameters without short circuit.

RT = 6kΩ…………….R1 = 2kΩ…………….R2 = 4kΩ

VT = 10V…………….. VR1 = 3.33V……… VR2 = 6.67V

IT = 1.67mA……….IR1 = 1.67mA……..IR2 = 1.67mA

PT = 16.7mW …… PR1 = 5.58mW …… PR2 = 11.16mW

Let’s find out what happens to the values ​​of this circuit when R1 is shorted.

VT = 10V

RT = 4kΩ………………R1 = 0…………………..R2 = 4kΩ

VT = 10V…………… VR1 = 0……. VR2 = 10V

IT = 2.5mA…………. IR1 = 0………………….. IR2 = 2.5 mA

PT = 25mW ………. PR1 = 0 ………………. PR2 = 25mW

Since there is only one resistance for current to flow through, the voltage is not divided.

Since R1 is shorted, there is no voltage drop (VR1 = 0).

Since R2 is the only resistor not bypassed by current, it maintains 5V (VR2 = 5V).

PR1 = VR1 x IR1

PR1 = 0 X 0

PR1 = 0

PR2 = VR2 x IR2

PR2 = 10V x 2.5mA

PR2 = 25mW

PRT = PR1 + PR2

PRT = 0 + 25mW

PRT = 25mW

So in summary, here’s what happened to the circuit when R1 was shorted out with a piece of metal wire.

1. Since the current travels the least path or resistance, it flowed through the metal instead of R1, making the resistance across R1=0.

2. Since the total resistance is lower, more current can flow through the circuit, making the current reading higher.

3. R1 = 0, makes the voltage drop across R1 also 0.

4. The overall voltage remains the same.

5. The power is higher because the current is higher.

Types of electrical circuits

A series connection with a voltage source (e.g. a battery or in this case a cell) and 3 resistance units

Two-pole components and electrical networks can be connected in series or in parallel. The resulting electrical network has two terminals and can itself participate in a series or parallel topology. Whether a two-terminal “object” is an electrical component (e.g. a resistor) or an electrical network (e.g. resistors in series) is a matter of perspective. In this article, “component” is used to refer to a two-port “object” that participates in the serial/parallel networks.

Components connected in series are connected along a single “electrical path” and each component has the same current equal to the current through the network. The voltage across the network is equal to the sum of the voltages across each component.[1][2]

Components connected in parallel are connected by multiple paths, and each component has the same voltage, which corresponds to the voltage on the network. The current through the network is equal to the sum of the currents through each component.

The previous two statements are equivalent except that the role of voltage and current is reversed.

A circuit consisting entirely of components connected in series is called a series circuit; A completely parallel circuit is also referred to as a parallel circuit. Many circuits can be analyzed as a combination of series and parallel circuits along with other configurations.

In a series circuit, the current flowing through each of the components is equal and the voltage across the circuit is the sum of the individual voltage drops across each component.[1] In a parallel circuit, the voltage across each of the components is the same and the total current is the sum of the currents flowing through each component.[1]

Imagine a very simple circuit consisting of four lightbulbs and a 12 volt car battery. When a wire connects the battery to a lightbulb, the next lightbulb, the next lightbulb, the next lightbulb, and then in an endless loop back to the battery, the lightbulbs are said to be connected in series. When each bulb is wired to the battery in a separate loop, the bulbs are said to be in parallel. If the four bulbs are connected in series, the same current will flow through all of them and the voltage drop will be 3 volts across each bulb, which may not be enough to light them. When the bulbs are connected in parallel the currents through the bulbs combine to form the current in the battery while the voltage drop across each bulb is 12 volts and they all light up.

In a series circuit, each device must function for the circuit to be complete. If an incandescent lamp burns out in a series circuit, the entire circuit is broken. In parallel circuits, each bulb has its own circuit, so all but one bulb could have burned out and the last one still work.

Series circuits [ edit ]

Series circuits are sometimes referred to as current coupled or daisy chain coupled. The electric current in a series circuit flows through all the components in the circuit. Therefore, all components in a series circuit carry the same current.

A series circuit has only one path through which its current can flow. Opening or breaking a series circuit at any point causes the entire circuit to “open” or stop operating. For example, if even one of the bulbs in an older style Christmas tree light string is burned out or removed, the entire string becomes inoperable until the bulb is replaced.

Current [edit]

i = i 1 = i 2 = ⋯ = i n {\displaystyle i=i_{1}=i_{2}=\cdots =i_{n))

In a series circuit, the current is the same for all elements.

tension [edit]

In a series connection, the voltage is the sum of the voltage drops of the individual components (units of resistance).

V = V 1 + V 2 + ⋯ + V n = I ( R 1 + R 2 + ⋯ + R n ) {\displaystyle V=V_{1}+V_{2}+\dots +V_{n}=I \left(R_{1}+R_{2}+\dots +R_{n}\right)}

Resistance Units[ edit ]

The total resistance of two or more resistors connected in series is equal to the sum of their individual resistances:

R total = R s = R 1 + R 2 + ⋯ + R n . {\displaystyle R_{\text{total}}=R_{\text{s}}=R_{1}+R_{2}+\cdots +R_{n}.}

Rs

Rs

Here, the subscript in denotes “series” and denotes resistance in a series.

Electrical conductivity is the reciprocal of resistance. The total conductivity of a series connection of pure resistors can therefore be calculated from the following expression:

1 G total = 1 G 1 + 1 G 2 + ⋯ + 1 G n . {\displaystyle {\frac {1}{G_{\text{total}}}}={\frac {1}{G_{1}}}+{\frac {1}{G_{2}}}+\ cdots +{\frac {1}{G_{n}}}.}

For a special case of two conductances connected in series, the total conductance is the same:

G total = G 1 G 2 G 1 + G 2 . {\displaystyle G_{\text{total}}={\frac {G_{1}G_{2}}{G_{1}+G_{2}}}.}

Inductors [ edit ]

Inductors follow the same law in that the total inductance of uncoupled inductors in series is equal to the sum of their individual inductances:

L t o t a l = L 1 + L 2 + ⋯ + L n {\displaystyle L_{\mathrm {total}}=L_{1}+L_{2}+\cdots +L_{n} }

In some situations, however, it is difficult to prevent neighboring inductors from affecting each other when a device’s magnetic field couples to its neighbors’ windings. This influence is defined by the mutual inductance M. For example, if two inductances are connected in series, there are two possible equivalent inductances, depending on how the magnetic fields of both inductances affect each other.

When there are more than two inductors, the mutual inductance between each of them and the way the coils affect each other complicates the calculation. For a larger number of coils, the combined total inductance is the sum of all mutual inductances between the various coils including the mutual inductance of any given coil with itself, which we call self-inductance or simply inductance. For three coils there are six mutual inductances M 12 {\displaystyle M_{12}} , M 13 {\displaystyle M_{13}} , M 23 {\displaystyle M_{23}} and M 21 {\displaystyle M_{21 }} , M 31 {\displaystyle M_{31}} and M 32 {\displaystyle M_{32}} . In addition, there are the three self-inductances of the three coils: M 11 {\displaystyle M_{11}} , M 22 {\displaystyle M_{22}} and M 33 {\displaystyle M_{33}} .

Because of this

L total = ( M 11 + M 22 + M 33 ) + ( M 12 + M 13 + M 23 ) + ( M 21 + M 31 + M 32 ) {\displaystyle L_{\text{total}}=\left( M_{11}+M_{22}+M_{33}\right)+\left(M_{12}+M_{13}+M_{23}\right)+\left(M_{21}+M_{31} }+M_{32}\right)}

By reciprocity, M i j {\displaystyle M_{ij}} = M.j. i {\displaystyle M_{ji}} so that the last two groups can be combined. The first three terms represent the sum of the self-inductances of the different coils. The formula can easily be extended to any number of series coils with mutual coupling. The method can be used to find the self-inductance of large coils of wire of any cross-sectional shape by calculating the sum of the mutual inductance of each turn of wire in the coil with every other turn, since in such a coil all turns are in series.

Capacitors [ edit ]

Capacitors follow the same law using the reciprocals. The total capacitance of capacitors in series is equal to the reciprocal of the sum of the reciprocals of their individual capacitances:

1 C total = 1 C 1 + 1 C 2 + ⋯ + 1 C n . {\displaystyle {\frac {1}{C_{\text{total}}}}={\frac {1}{C_{1}}}+{\frac {1}{C_{2}}}+\ cdots +{\frac {1}{C_{n}}}.}

switch [ edit ]

Two or more switches in series form a logical AND; the circuit carries current only when all switches are closed. See AND gate.

Cells and batteries[ edit ]

A battery is a collection of electrochemical cells. When the cells are connected in series, the voltage of the battery is the sum of the cell voltages. For example, a 12-volt car battery contains six 2-volt cells connected in series. Some vehicles such as B. Trucks have two 12 volt batteries in series to power the 24 volt system.

Parallel circuits[ edit ]

Comparison of effective resistance, inductance and capacitance of two resistors, inductors and capacitors in series and in parallel

“Parallel” redirects here. For the 2017 album by Dhani Harrison, see In Parallel (album)

When two or more components are connected in parallel, they have the same potential difference (voltage) at their ends. The potential differences between the components are equal, and they also have identical polarities. The same voltage is applied to all circuit parts connected in parallel. The total current is the sum of the currents through the individual components according to Kirchhoff’s current law.

tension [edit]

In a parallel circuit, the voltage is the same for all elements.

V = V 1 = V 2 = ⋯ = V n {\displaystyle V=V_{1}=V_{2}=\dots =V_{n}}

Current [edit]

The current in each individual resistor is determined by Ohm’s law. Calculating the voltage results

i total = i 1 + i 2 + ⋯ + i n = V ( 1 R 1 + 1 R 2 + ⋯ + 1 R n ) . {\displaystyle I_{\text{total}}=I_{1}+I_{2}+\cdots +I_{n}=V\left({\frac {1}{R_{1}}}+{\ frac {1}{R_{2}}}+\cdots +{\frac {1}{R_{n}}}\right).}

Resistance Units[ edit ]

To find the total resistance of all components, add the reciprocals of the resistances of each component and take the reciprocal of the sum. The total resistance is always less than the value of the smallest resistance:

1 R total = 1 R 1 + 1 R 2 + ⋯ + 1 R n . {\displaystyle {\frac {1}{R_{\text{total}}}}={\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}+\ cdots +{\frac {1}{R_{n}}}.}

For only two resistances, the unrequited expression is fairly simple:

R total = R 1 R 2 R 1 + R 2 . {\displaystyle R_{\text{total}}={\frac {R_{1}R_{2}}{R_{1}+R_{2}}}.}

This sometimes goes through the mnemonic product over the sum.

For N equal resistors in parallel, the reciprocal sum expression simplifies to:

1 row total = N 1 row. {\displaystyle {\frac {1}{R_{\text{total}}}}=N{\frac {1}{R}}.}

R total = R N . {\displaystyle R_{\text{total}}={\frac {R}{N}}.}

and therefore to:

R i {\displaystyle R_{i}} To find the current in a resistive component, use Ohm’s law again:

i i = V R i . {\displaystyle I_{i}={\frac {V}{R_{i}}}\,.}

The components divide the current according to their reciprocal resistances, i.e. with two resistances

i 1 i 2 = R 2 R 1 . {\displaystyle {\frac {I_{1}}{I_{2}}}={\frac {R_{2}}{R_{1}}}.}

An old term for devices connected in parallel is multiple, such as multiple connections for arc lamps.

Since the electrical conductivity is reciprocal to the resistance, the expression for the total conductivity of a parallel circuit of resistors is:

G total = G 1 + G 2 + ⋯ + G n . {\displaystyle G_{\text{total}}=G_{1}+G_{2}+\cdots +G_{n}.}

The relationships for total conductance and resistance are complementary: the expression for a series connection of resistances is the same as for a parallel connection of conductances and vice versa.

Inductors [ edit ]

Inductors follow the same law in that the total inductance of uncoupled inductors connected in parallel is equal to the reciprocal of the sum of the reciprocals of their individual inductors:

1 L total = 1 L 1 + 1 L 2 + ⋯ + 1 L n . {\displaystyle {\frac {1}{L_{\text{total}}}}={\frac {1}{L_{1}}}+{\frac {1}{L_{2}}}+\ cdots +{\frac {1}{L_{n}}}.}

If the inductors are in each other’s magnetic fields, this approach is invalid due to mutual inductance. If the mutual inductance between two coils connected in parallel is M, the equivalent inductance is:

1 L total = L 1 + L 2 − 2 M L 1 L 2 − M 2 {\displaystyle {\frac {1}{L_{\text{total}}}}={\frac {L_{1}+L_{ 2}-2M}{L_{1}L_{2}-M^{2}}}}

If L 1 = L 2 {\displaystyle L_{1}=L_{2}}

L total = L + M 2 {\displaystyle L_{\text{total}}={\frac {L+M}{2}}}

The sign of M {\displaystyle M} depends on how the magnetic fields affect each other. For two equally closely coupled coils, the total inductance is close to that of each individual coil. If the polarity of a coil is reversed so that M is negative, then the parallel inductance is nearly zero, or the combination is nearly non-inductive. M is assumed to be almost equal to L in the “tightly coupled” case. However, if the inductances are not equal and the coils are closely coupled, near short circuit conditions and high circulating currents M can occur at both positive and negative values, which can cause problems.

More than three inductors become more complex and the mutual inductance of each inductor to each other inductor and their mutual influence must be considered. For three coils there are three mutual inductances M 12 {\displaystyle M_{12}} , M 13 {\displaystyle M_{13}} and M 23 {\displaystyle M_{23}} . This is best handled by matrix methods and summing the terms of the inverse of the matrix L {\displaystyle L} (in this case 3 × 3).

The associated equations have the form:

v i = ∑ j L i , j d i j d t {\displaystyle v_{i}=\sum _{j}L_{i,j}{\frac {di_{j)){dt)))

Capacitors [ edit ]

The total capacitance of capacitors connected in parallel is equal to the sum of their individual capacitances:

C total = C 1 + C 2 + ⋯ + C n . {\displaystyle C_{\text{total}}=C_{1}+C_{2}+\cdots +C_{n}.}

The working voltage of a parallel connection of capacitors is always limited by the smallest working voltage of a single capacitor.

switch [ edit ]

Two or more switches in parallel form a logical OR; the circuit is live when at least one switch is closed. See OR gate.

Cells and batteries[ edit ]

When the cells of a battery are connected in parallel, the battery voltage is equal to the cell voltage, but the current supplied by each cell is a fraction of the total current. For example, if a battery consists of four identical cells connected in parallel and supplies a current of 1 ampere, the current supplied by each cell is 0.25 amperes. If the cells are not identical, cells with higher voltages will try to charge those with lower ones, potentially damaging them.

Batteries connected in parallel were often used to power the valve filaments in portable radios. Lithium-ion batteries (especially laptop batteries) are often connected in parallel to increase the ampere-hour rating. Some solar power systems have batteries connected in parallel to increase storage capacity; A good approximation of total amp hours is the sum of all amp hours from batteries connected in parallel.

Combining guide values[edit]

From Kirchhoff’s circular laws we can derive the rules for the combination of conductances. For two conductances G 1 {\displaystyle G_{1}} and G 2 {\displaystyle G_{2}} connected in parallel, the voltage across them is the same and, according to Kirchhoff’s current law (KCL), the total current

I eq = I 1 + I 2 . {\displaystyle I_{\text{eq}}=I_{1}+I_{2}.}

Substituting Ohm’s law for conductance gives:

G eq V = G 1 V + G 2 V {\displaystyle G_{\text{eq}}V=G_{1}V+G_{2}V}

G eq = G 1 + G 2 . {\displaystyle G_{\text{eq}}=G_{1}+G_{2}.}

and the equivalent conductivity will be,

For two series connected conductances G 1 {\displaystyle G_{1}} and G 2 {\displaystyle G_{2}} the current through them is equal and Kirchhoff’s voltage law tells us that the voltage across them is the sum of the two Voltages above each conductance, that is,

V eq = V 1 + V 2 . {\displaystyle V_{\text{eq}}=V_{1}+V_{2}.}

Substituting Ohm’s law for conductivity then gives:

I G eq = I G 1 + I G 2 {\displaystyle {\frac {I}{G_{\text{eq}}}}={\frac {I}{G_{1}}}+{\frac { I}{ G_{2}}}}

1 G eq = 1 G 1 + 1 G 2 . {\displaystyle {\frac {1}{G_{\text{eq}}}}={\frac {1}{G_{1}}}+{\frac {1}{G_{2}}}.}

which in turn gives the formula for the equivalent conductance,

This equation can easily be rearranged, although this is a special case that is only rearranged for two components.

G eq = G 1 G 2 G 1 + G 2 . {\displaystyle G_{\text{eq}}={\frac {G_{1}G_{2}}{G_{1}+G_{2}}}.}

G eq = G 1 G 2 G 3 G 1 G 2 + G 1 G 3 + G 2 G 3 . {\displaystyle G_{\text{eq}}={\frac {G_{1}G_{2}G_{3}}{G_{1}G_{2}+G_{1}G_{3}+G_{ 2}G_{3}}}.}

Spelling [ edit ]

The following applies to three conductance values ​​connected in series:

The value of two parallel components in equations is often represented by the parallel operator, two vertical lines (∥), using the notation for parallel lines borrowed from geometry.

R e q ≡ R 1 ∥ R 2 ≡ ( R 1 − 1 + R 2 − 1 ) − 1 ≡ R 1 R 2 R 1 + R 2 {\displaystyle R_{\ mathrm {eq}} \ equiv R_{1} \ parallel R_{2}\equiv \left(R_{1}^{-1}+R_{2}^{-1}\right)^{-1}\ equiv {\frac {R_{1}R_{2 }}{R_{1}+R_{2}}}}

This simplifies expressions that would otherwise be complicated by expanding the terms. For example:

R 1 ∥ R 2 ∥ R 3 ≡ R 1 R 2 R 3 R 1 R 2 + R 1 R 3 + R 2 R 3 . {\displaystyle R_{1}\parallel R_{2}\parallel R_{3}\equiv {\frac {R_{1}R_{2}R_{3}}{R_{1}R_{2}+R_{ 1}R_{3}+R_{2}R_{3}}}.}

If n components are parallel, then

R eq = ( ∑ i n R i − 1 ) − 1 {\displaystyle R_{\text{eq}}=\left(\sum _{i}^{n}{R_{i}}^{-1} \ right)^{-1}}

Applications[edit]

A common application of series connection in consumer electronics is batteries, which use multiple cells connected in series to obtain a suitable operating voltage. Two disposable zinc cells connected in series could power a flashlight or remote control at 3 volts; The battery pack for a hand-held power tool may contain a dozen lithium-ion cells connected in series to provide 48 volts.

Series circuits were previously used for lighting in electric multiple units. For example, if the supply voltage is 600 volts, eight 70 volt bulbs can be connected in series (560 volts total) plus a resistor to drop the remaining 40 volts. Series circuits for train lighting were replaced, first by motor-generators, then by solid-state devices.

Series resistance can also be applied to the arrangement of blood vessels within a given organ. Each organ is supplied by a major artery, smaller arteries, arterioles, capillaries and veins arranged in series. The total resistance is the sum of the individual resistances as expressed by the following equation: R total = R artery + R arterioles + R capillaries. The arterioles contribute most of the resistance in this series.[3]

Parallel resistance is illustrated by the circulatory system. Each organ is supplied by an artery that branches off the aorta. The total resistance of this parallel arrangement is expressed by the following equation: 1/R total = 1/R a + 1/R b + … + 1/R n . R a , R b and R n are the resistances of the renal, hepatic and other arteries, respectively. The total resistance is less than the resistance of any of the individual arteries.[3]

See also[edit]

References[edit]

In order to determine the amount or magnitude of electric current flowing through an electrical or electronic circuit, we must apply certain laws or rules that allow us to write these currents in the form of an equation. The network equations used are in accordance with Kirchhoff’s Laws and since we are dealing with electrical circuits we will consider current Kirchhoff’s Law (KCL).

Gustav Kirchhoff’s current law is one of the fundamental laws used for circuit analysis. His current law states that for a parallel path, the total current entering a circuit connection is exactly equal to the total current leaving the same connection. This is because there is no other place it can go as no charge is lost.

In other words, the algebraic sum of ALL currents entering and leaving a junction must equal zero, as follows: Σ I IN = Σ I OUT .

This idea of ​​Kirchhoff’s is commonly known as conservation of charge, since the current is conserved around the junction with no loss of current. Let’s look at a simple example of Kirchhoff’s current law (KCL) when applied to a single intersection.

A single crossing

Here in this simple example with a single node, the current I T leaving the node is the algebraic sum of the two currents I 1 and I 2 entering the same node. That is IT = I 1 + I 2 .

Note that we could also correctly write this as the algebraic sum of: I T – (I 1 + I 2 ) = 0.

So if I 1 equals 3 amps and I 2 equals 2 amps then the total current I T leaving the junction is 3 + 2 = 5 amps and we can use this basic law for any number of junctions or nodes The sum of the incoming and outgoing streams is the same.

Even if we reversed the directions of the currents, the resulting equations would still hold for I 1 or I 2 . As I 1 = I T – I 2 = 5 – 2 = 3 amps and I 2 = I T – I 1 = 5 – 3 = 2 amps. Therefore, we can think of the currents entering the junction as positive (+) while those leaving the junction are negative (-).

Then we can see that the mathematical sum of the currents either entering or leaving the junction, and in whichever direction, is always zero, and this forms the basis of Kirchhoff’s Junction Rule, better known as Kirchhoff’s Current Law or ( KCL). .

resistors in parallel

Let’s look at how we could apply Kirchhoff’s current law to parallel resistors, regardless of whether the resistances in those branches are equal or unequal. Consider the following schematic:

In this simple parallel resistance example, there are two different power connections. Node one occurs at node B, node two occurs at node E. Thus we can use Kirchhoff’s node rule for the electric currents at these two distinct nodes, for the currents entering the node and for the currents leaving the node.

Initially all current I T leaves the 24 volt supply and arrives at point A and from there enters node B. Node B is a node because the current can now split in two different directions, with part of the current flowing down through resistor R1, with the remainder continuing through resistor R2 via node C. Note that the currents flowing into and out of a node are commonly called branch currents.

We can use Ohm’s Law to determine the individual branch currents through each resistor as: I = V/R, so:

For current branch B to E through resistor R 1

For current branch C to D through resistor R 2

From above we know that Kirchhoff’s current law states that the sum of currents entering an intersection must equal the sum of currents exiting the intersection, and in our simple example above there is one current, IT , which enters the junction at node B and flows two currents leaving the junction, I 1 and I 2 .

Now that we know from the calculation that the currents leaving the junction at node B, I 1 equal 3 amps and I 2 equal 2 amps, the sum of the currents entering the junction at node B must equal 3 + 2 = 5 amps. Thus Σ IN = I T = 5 amps.

In our example we have two different nodes at node B and node E, so we can confirm this value for IT as the two streams recombine at node E again. So for Kirchhoff’s node rule to apply, the sum of the currents at point F must equal the sum of the currents flowing out of the junction at node E.

Since the two currents entering node E are 3 amps and 2 amps respectively, the sum of the currents entering point F is therefore: 3 + 2 = 5 amps. Thus Σ IN = I T = 5 amperes and therefore Kirchhoff’s current law applies since this is the same value as the current output point A.

Applying KCL to more complex circuits.

We can use current Kirchhoff’s law to find the currents that flow in more complex circuits. Hopefully we now know that the algebraic sum of all currents at a node (connection point) is zero, and with this idea in mind it is a simple case to determine the currents entering and exiting a node. Consider the following circuit.

Kirchhoff’s current law example No. 1

In this example there are four distinct nodes for the current to either separate or merge at nodes A, C, E and node F. The supply current I T separates at node A, flows through resistors R1 and R2, and rejoins at node C before separating through resistors R3, R4, and R5 and finally recombination again at node F.

But before we can calculate the individual currents flowing through each resistive branch, we must first calculate the total current of the circuit, I T . Ohm’s law tells us that I = V/R and since we know the value of V, 132 volts we need to calculate the circuit resistances as follows.

circuit resistance R AC

Thus, the equivalent circuit resistance between nodes A and C is calculated to be 1 ohm.

Circuit resistance R CF

Thus, the equivalent circuit resistance between nodes C and F is calculated as 10 ohms. Then the total circuit current I T is given as follows:

Give us an equivalent circuit of:

Kirchhoff’s Current Law Equivalent Circuit

Hence V = 132V, R AC = 1Ω, R CF = 10Ω and I T = 12A.

Having determined the equivalent parallel resistances and the supply current, we can now calculate the individual branch currents and confirm them using Kirchhoff’s connection rule as follows.

Thus I 1 = 5 A, I 2 = 7 A, I 3 = 2 A, I 4 = 6 A and I 5 = 4 A.

We can confirm that Kirchoff’s current law applies around the circuit, using node C as a reference point to calculate the currents entering and leaving the connection as:

We can also verify that Kirchhoff’s current law holds, since the currents entering the junction are positive while those leaving the junction are negative, so the algebraic sum is: I 1 + I 2 – I 3 – I 4 – I 5 = 0, which corresponds to 5 + 7 – 2 – 6 – 4 = 0.

So we can confirm through analysis that Kirchhoff’s current law (KCL), which states that the algebraic sum of the currents at a node in a circuit network is always zero, is true and correct in this example.

Kirchhoff’s Actual Law Example #2

Find the currents flowing in the following circuit using only Kirchhoff’s current law.

I T is the total current flowing through the circuit driven by the 12V supply voltage. At point A, I 1 equals IT, so there is an I 1 *R voltage drop across resistor R 1 .

The circuit has 2 branches, 3 nodes (B, C and D) and 2 independent loops, therefore the I*R voltage drops around the two loops are:

Loop ABC ⇒ 12 = 4I 1 + 6I 2

Loop ABD ⇒ 12 = 4I 1 + 12I 3

Since Kirchhoff’s current law states that at node B I 1 = I 2 + I 3 , we can therefore replace the current I 1 by (I 2 + I 3 ) in the two loop equations below and then simplify.

Kirchhoff loop equations

We now have two simultaneous equations related to the currents flowing around the circuit.

Eq. No. 1 : 12 = 10I 2 + 4I 3

Eq. #2 : 12 = 4I 2 + 16I 3

By multiplying the first equation (Loop ABC) by 4 and subtracting Loop ABD from Loop ABC, we can reduce both equations to give us the values ​​of I 2 and I 3

Eq. No. 1 : 12 = 10I 2 + 4I 3 ( x4 ) ⇒ 48 = 40I 2 + 16I 3

Eq. #2 : 12 = 4I 2 + 16I 3 ( x1 ) ⇒ ​​12 = 4I 2 + 16I 3

Eq. #1 – Eq. No 2 ⇒ 36 = 36I 2 + 0

Substituting I 2 for I 3 gives the value of I 2 as 1.0 ampere

Now we can use the same procedure to find the value of I 3 by multiplying the first equation (loop ABC) by 4 and the second equation (loop ABD) by 10. Again, we can reduce both by subtracting loop ABC from loop ABD equations to give us the values ​​of I 2 and I 3

Eq. No. 1 : 12 = 10I 2 + 4I 3 ( x4 ) ⇒ 48 = 40I 2 + 16I 3

Eq. No. 2 : 12 = 4I 2 + 16I 3 ( x10 ) ⇒ 120 = 40I 2 + 160I 3

Eq. #2 – Eq. No 1 ⇒ 72 = 0 + 144I 3

Thus substituting I 3 for I 2 gives the value of I 3 as 0.5 amps

As Kirchhoff’s node rule states: I 1 = I 2 + I 3

The supply current flowing through resistor R 1 is given as: 1.0 + 0.5 = 1.5 amps

Thus I 1 = I T = 1.5 amps, I 2 = 1.0 amps and I 3 = 0.5 amps and from this information we could calculate the I*R voltage drops across the devices and at the various points (nodes ) around the circuit.

We could have easily and simply solved the circuit of example two using only Ohm’s law, but we’ve used current Kirchhoff’s law here to show how it’s possible to solve more complex circuits if we don’t just apply Ohm’s law be able.

Electrical resistance formula

Electrical resistance is a quantity that measures the resistance that a device or material offers to the flow of electricity. The SI unit of resistance is the ohm (Ω). A resistor is an electrical component used to provide the desired resistance in an electrical circuit. In this article we will discuss the various electrical resistance formulas used to find the resistance a conductor offers to the flow of electricity.

Formula for electrical resistance

If we know the length and cross-sectional area of ​​a conductor, the electrical resistance of a conductor is the product of the conductor’s resistivity and the conductor length divided by the cross-sectional area of ​​the conductor. Mathematically it is represented as follows:

\(\begin{array}{l}R=\rho \frac{l}{A}\end{array} \)

Where,

R is the resistance

\(\begin{array}{l}\rho\end{array} \) is the specific resistance of the conductor

is the resistivity of the conductor l is the length of the conductor

A is the area of ​​the cross section of the conductor

From the above formula it can be seen that resistance is directly proportional to the length of the conductor and inversely proportional to the area of ​​the conductor. This formula can be better understood using a hookah analogy as follows:

If the pipe is longer, the length is greater and therefore the resistance to water flow is high.

When the pipe is wider, the area of ​​the pipe is larger and therefore the resistance to water flow is less.

Resistance calculation according to Ohm’s law

The electrical resistance of a conductor can be calculated using Ohm’s law if the current and the voltage drop across it are known. The formula for calculating resistance using Ohm’s law is as follows:

\(\begin{array}{l}R=\frac{V}{I}\end{array} \)

Where,

R is the resistance of the resistor R in ohms (Ω)

is the resistance of the resistor R in ohms (Ω) V is the voltage drop across the resistor in volts

is the voltage drop across the resistor in volts I is the current flowing through the resistor (A)

Electrical resistance problems

Example 1: In a circuit, a current of 6.00 A flows through a resistor. The voltage drop from one end of the resistor to the other is 150V. What is the value of the resistor?

Solution:

Here we know the current and the voltage drop across the conductor, so we can use Ohm’s law to find the resistance as follows:

\(\begin{array}{l}R=\frac{V}{I}\end{array} \)

Substituting the values ​​into the above equation, we get

\(\begin{array}{l}R=\frac{150\,V}{6\,A}=25\,\Omega\end{array} \)

The resistance of the resistor in the circuit is

\(\begin{array}{l}25\,\Omega\end{array} \)

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