Statement : “Algebraic sum of branch currents at node is zero at all instance of time.”

Or

“At any node (junction) in a network, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node.”

Fig.1 Electrical Network with 5 branches

To understand the statement, let’s consider one example. As shown in Fig. 1, the network has one node N and 5 branches. The current in five branches is I_1, I_2, I_3, I_4, and I₅. The current I₁, I₂, and I₃ are entering the node N, while current I₄ and I₅ are leaving the node.

Sign Convention

If current is entering the node, then consider it as positive current. Similarly, if current is leaving the node then it can be considered as negative current. The same is shown in Fig.2.

Fig.2 Sign Convention for Kirchhoff’s Current Law (KCL)

Therefore, for the network shown in Fig. 1, current I₁, I₂, and I₃ will be positive, while current I₄ and I₅ will be negative.

And according the Kirchhoff’s Current Law, the algebraic sum of all these current is zero.

Therefore, I₁ + I₂ + I₃ – I₄ – I₅ = 0

∴ I₁ + I₂ + I₃ = I₄ + I₅ ——-(1)

It shows that, the sum of currents entering the node is equal to the sum of currents leaving the node.

KCL is a law of Conservation of Charge

The current is the rate at which the charge is flowing.

∴ Current (I) = Q/t

Therefore, the equation 1 can be written as

And further after the simplification, Q₁ + Q₂ + Q₃ = Q₄ + Q₅ ———-(2)

Equation 2 shows that, the charge which is entering the node is equal to the charge which is leaving the node.

Therefore, Kirchhoff’s Current Law (KCL) is the law of conservation of charge.

Example

For the given circuit if, I₂ = 2A, I₄ = -1A and I₅ = -4A then find current I₆

Fig. 3 Kirchhoff’s Current Law (KCL) Example

Solution:

Applying KCL at node B,

I₃ + I₆ = I1 and I₁ = 2A (Given)

∴ I₆ = 2 – I₃ ——— (3)

Similarly, applying KCL at node C,

I₂ + I₅ = I₃ and I₅ = -4 A (Given)

∴ I₃ = I₂ – 4 ———-(4)

Similarly, applying KCL at node A,

I₁ + I₄ = I₂ and I₄ = -1 A and I₁ = 2A (Given)

∴ I₂ = 2 -1 = 1A

From equation 4, I₃ = 1 – 4 = -3A

And by putting the value of I₃ in equation 3,

I₆ = 2 – (-3) = 5A

Therefore, for the given circuit, current I₆ = 5A

For more information on Kirchhoff’s Current Law (KCL) check this video:

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Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL) are very fundamental laws in the electrical circuit. Using these laws, we can find the voltage and current in the electrical circuit.

Statement: The algebraic sum of all the branch voltages around any closed loop in the network or circuit is zero at all instant of time.

Let’s understand the statement through one example.

Fig. 1 Kirchhoff’s Voltage Law in the Electrical Circuit

Fig.1 shows the electrical circuit which consists of one voltage source and three passive circuit elements. The passive circuit elements could be a resistor, capacitor, or inductor. But in general form, here they are represented as Z₁, Z_2, and Z₃. Let’s say the voltages across these elements are V₁, V_2, and V₃ respectively.

According to the Kirchhoff’s Voltage Law, if we mover around any electrical circuit (either clockwise or anti-clockwise) and add the voltages drop across each element then the algebraic sum of all the voltages will be zero.

Let’s say as shown in Fig.2, we are moving in the clockwise direction, starting from point A.

Fig.2 Sign convention for Kirchhoff’s Voltage Law (KVL)

While moving in the clockwise direction, the first element that we will come across is the voltage source Vs. And while we are moving, we are moving from the negative to the positive potential. So, we will use the following sign convention for the KVL.

Sign Convention for Kirchhoff’s Voltage Law (KVL)

Whenever we are moving from the positive (+) to the negative (-) terminal across any element or in other words, if there is a drop in the potential across the element then we can consider that voltage as the negative voltage.

Similarly, whenever we are moving from the negative (-) terminal to the positive (+) terminal across any element or in other words, if there is a rise in the potential across the element then we can consider that voltage as the positive voltage. The same is shown in Fig.3.

Fig. 3 Sign Conventions for the Kirchhoff’s Voltage Law

As shown in Fig.2, starting from point A, when we move in the clockwise direction then the first element is the voltage source Vs. And while moving, since there is a rise in the potential, we can consider that voltage as a positive voltage. Similarly, when we are moving from point B to point C, there is a voltage drop across the element Z₁. That means the voltage V₁ can be considered as the negative voltage. Similarly, while moving across the element Z₂ and Z₃, there is a drop across each element. That means the voltage across element Z₂ and Z₃ (V₂ and V₃) will be negative.

According to the Kirchhoff’s Voltage Law, the algebraic sum of all these voltage is zero.

That means Vs + ( – V1) + (- V₂ ) + ( – V₃ ) = 0

=> Vs – V₁ – V₂ – V₃ = 0

=> Vs = V₁ + V₂ + V₃

KVL is the law of Conservation of Energy

Kirchhoff’s Voltage Law is the low of conservation of energy. Let’s prove it. The voltage V can also be written as

In the above case, V_S = V₁ + V₂ + V₃ can be written as

That means E_S= E₁ + E₂ + E₃

Or it can be said that the energy supplied by the voltage source is equal to the energy dissipated across three elements. That means KVL is the law of conservation of energy.

Example

Find the current I and the voltage across 15 Ω resistor.

Solution:

First, let’s denote the voltages across each element. Let’s say the voltage drop across 5 Ω, 10 Ω, and 15 Ω resistor are V1, V2, and V3 respectively.

Applying Kirchhoff’s voltage law,

5V – V₁ – V₂ – 2V – V₃ = 0

Since, moving in the clockwise direction, there is a rise in potential only across 5V voltage source and there is a drop in the potential across remaining elements.

That means, 3V = V₁ + V₂ + V₃ ——- (1)

Since all the elements are connected in series, the current flowing through each element is the same. Let’s say the current in the circuit is I.

Therefore, using Ohm’s law (V= I x R), V₁ = 5 x I, V₂ = 10 x I and V₃ = 15 x I

From the above equation 1,

3V = (5 x I ) + ( 10 x I ) + (15 x I)

=> 3V = 30 x I

=> I = 0.1 A

And the voltage across 15 Ω resistor (V₃) = 15 x I = 15 V x 0.1A = 1.5 V

From the above example, we can see that, using the Kirchhoff’s voltage law, it is possible to find the current and the voltage across any element in the electrical circuit.

For more information, you can refer this video tutorial on KVL,

The post Kirchhoff’s Voltage Law (KVL) Explained appeared first on ALL ABOUT ELECTRONICS.

Resistors in Series

As shown above, when the n- resistors are connected in series connection, then the total resistance will be the summation of individual resistance.

Proof : 

Let’s say, three resistors are connected in series with one voltage source V. And current flowing through the circuit is I.

If V1, V2 and V3 are the voltage drop across each element then applying the KVL in the loop,

V – V1 – V2 – V3 = 0

⇒  V = V1 + V2 + V3

⇒ V = (I x R1 + I x R2 + I x R3)

⇒ V = I (R1 + R2 + R3) —— (1)

Let’s say, Req is the equivalent resistance of the three resistors.

If three resistors are replaced by the equivalent resistance, then the equivalent circuit can be represented as follows:

Applying the  Kirchhoff’s Voltage Law (KVL) for the equivalent circuit,

V – (I x Req ) = 0

⇒ V = I x Req —— (2)

Equating the equation (1) and (2),

Req = R1 + R2 +R3

Or in general, when n- number of resistors are connected in series then the total or equivalent resistance can be given as

Req = R1 + R2 + R3 + —– + Rn

Resistors in Parallel :

The resistors are said to be connected in parallel when their terminals are connected to the same node.

For example, as shown in the figure, the three resistors are connected in the parallel, as one of their terminals is connected to the node A, while the second terminals are connected to the node B.

If n- resistors are connected in parallel then the equivalent resistance can be given by the following expression.

Proof:

As shown in the figure, three resistors are connected in parallel with one voltage source V.

Here, the current through each resistor will be different. But as they are connected in parallel, the voltage across each resistor will be the same.

If V1, V2 and V3 is the voltage across the resistor R1, R2 and R3 respecitvely then

V = V1 = V2 = V3 ——(1)

If the I is the total current supplied by the voltage source, then applying the KCL at the top node,

I = I1 +I2 + I3 ——- (2)

Now, according to the Ohm’s law,

I1 = V1/ R1

I2 = V2 / R2

I3 = V3 / R3

Using, equation (1) and (2)

I = V x  { (1 / R1) + (1/ R2) + (1/ R3) } ——— (3)

If three resistors are replaced by the equivalent resistance then the equivalent circuit can be represented as follows:

∴ V = I x Req

⇒ I = V / Req  ———(4)

Comparing equation (3) and (4)

1 / Req = { (1 / R1) + (1/ R2) + (1/ R3) }

Or, in general, when n- resistors are connected in parallel, then equivalent resistance Req can be given by the following expression.

1 / Req = { (1 / R1) + (1/ R2) + (1/ R3) + —–  + (1/ Rn)}

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A branch represents the single circuit elements like resistor, capacitor, inductor, voltage, or current source.

                                                                  Figure. 1

For example, for the circuit shown in figure 1, there are five branches. A 10 V voltage source, 2A current source, 4 Ω, 5 Ω, and 3 Ω resistors.

What is Node?

A node is a point in the circuit where two or more circuit elements (or branches) are connected.                               Figure. 2 ( Representation of Four Nodes in the Circuit)

For example, as shown in Figure 2, the above circuit contains the Four nodes. The node A, B, C, and D.

What is Loop?

Any closed path in the circuit is called as a loop.

A loop is a closed path formed by starting at a node, passing through a
set of nodes, and returning to the starting node without passing through
any node more than once.

Figure. 3 (Representation of Loop in the Circuit)

For example, as shown in Figure.3, the circuit contains three loops.

The first is loop A-B-D-A, the second loop is B-C-D-B. And the third loop is A-B-C-D-A.

What is Mesh?

A mesh is a closed path in the circuit, which does not contain any other close path inside it.

For example, as shown in Figure.3, loop 1(A-B-D-A) and loop 2 (B-C-D-B) does not contain any other closed path within them. And they are the example of the Mesh. While loop 3 (A-B-C-D-A) contains loop 1 and loop 2 within it. So, it can’t be called as a Mesh.

Note: All  Mesh are loops but not all the loops are Mesh.

For more information, check this video:

The Concept of Loop, Mesh, Node and Branch

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