Kirchoff’s Circuit Law

Kirchoff’s Circuit Law

There are some circuits we can not find the current or voltage moving within the circuit simply by using Ohm’s law. For this type of circuit (complex circuit), we use specific law or rule to get the circuit equation which in turns will lead to obtaining the current or voltage within a circuit. Kirchoff’s Circuit Law is used for this purpose. A German physicist in 1845 known as Gustav Kirchoff developed a set of laws which deal with the conservation of current and energy within electrical circuits. These two laws are commonly known as: Kirchoffs Circuit Laws which are Kirchoff’s Current Law (KCL) and Kirchoff’s Voltage Law (KVL).

 

Kirchoff’s First Law or Kirchoff’s Current Law (KCL).

Kirchoff’s Current Law states that the “ total current entering a junction or node is equal to the current leaving the node in a closed circuit. It can also be stated as the algebraic sum of ALL the currents entering and leaving a node must be equal to zero.Kirchhoff’s  current law is a statement about the conservation of charge flow through a circuit.

Kirchoff's Current Lsw

 

From the diagram, there are five lines representing the flow of current. Let the current entering the node be positive and the one leaving the node be negative or vice versa.

So we have :

Current entering the node = I2 + I5

Current leaving the node=  -(I1+ I3 + I4)

Putting the above together, we now have

I2 + I5 – I1 – I3 – I4 =0

 

Kirchoff’s Second Law or Kirchoff’s Voltage Law (KVL)

Kirchoff’s Voltage Law states that “ in any closed circuit, the total voltage around the loop is equal to the sum of all the voltage drops within the same the loop” which is also equal to zero. It can also be stated as the algebraic sum of all voltages within the loop must be equal to zero. Kirchoff Voltage Law is also known as the Conservation of Energy .

 

Common Circuit Theory Terms:

Circuit – A circuit is the pathway through which electrical current flows.Path – A path is a single line of connecting elements or sources.

Node – A Node any point on a circuit where two or more circuit elements intersect. A node is indicated by a dot.

Branch – A branch is a single or group of components  such as voltage source or a current source or a resistor which are connected between two nodes.

Loop – A loop is any closed path in a circuit.

 

 Steps In Solving Problems On KCL And KVL

1. Determine the direction of current whenever solving circuits via Kirchhoff’s laws. The direction of current can be supposed through clockwise or anticlockwise direction. You can choose the direction of the current whether clockwise or anticllockwise, whichever you choose to use, maintain the same direction till the final solution or equations of the circuit.

Note: The direction of current by the battery in clockwise direction is taken as positive (+) while the one in anticlockwise direction is taken as negative (-).

2. Derive the KCL equation for the circuit in each node.

3. Derive the KVL equation for the circuit in each loop

4. Solve the simultaneous equation in order to get the unknown currents.

 

Kirchoff’s Circuit Law Solved Example

Kirchoff's Circuit Law Example

Find the current and voltage in the 40 Ohms resistor.

 

 

 

Step1

Kirchoff's Circuit Law Problem

This circuit  has 5 branches, 2 nodes (C & A), 2 independent loops (loop 1 & 2) and 1 dependent loop (loop 3).

 

 

 

Step2

Current enters the node C from junction BC (I1) and CD (I2) and leaves the node through CA ( I3).

Applying KCL: l1+ I2 – I3=0 or  I3=I1 + I2……………(1)

 

Step 3:

Applying KVL

From Loop 1: 12=R1 I1 +R3 I3; 12I1 +40I3=12……..(2)

From Loop 2: 24=R2 I2 +R3 I3; 6I2 + 40I3=24………(3)

From Loop 3:12-24=I1 R1-R2 I2; 12I1- 6 I2= -12……(4)

 

Step 4:

Solve simultaneously by taking any two equations from step 3.

12I1 +40I3=12

6I2 + 40I3=24

The above has to be reduced to two unknowns in order to solve simultaneously. That can be achieved by simply substituting equation 1 (I3=I1 +I2) into both equations above.

So the equations becomes:

42I1+ 40I2=12

40I1+ 46I2=24

Solving simultaneously, I1= -1.22, I2=1.59

Therefore I3=-1.22+1.59=0.36

The voltage in R3 will be V3=0.36*40=14.4 (Ohm’s law V=IR).

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