A Simple DC Circuit Kirchhoff's - KVL and KCL Illustration

Kirchhoff's Current and Voltage Laws Illustrated

A Simple DC Circuit

                                                                      Fig.1

This is a simple DC Circuit as one may see, having only resistors and independent sources of DC voltages of  ±10V and ±5V DC. This "±" symbol means that looking from one side the voltage is + 10V and from the other side it's -10V and the same for the ±5V DC Source. 

Now the allocation of sign during analysis of the circuit is very important, since according to Kirchhoff's Laws, the algebraic sum hold primary importance, in which the "sign" is very important.

Kirchhoff's Current Law (KCL):

This states that the algebraic summation of the currents, either incoming or outgoing, from a node (having at least two branches) or a junction (having more than 2 branches) is equal to Zero.

Kirchhoff's Voltage Law (KVL):

This states that the algebraic summation of the voltage drops and voltage sources in any given closed loop/mesh is equal to Zero.

Application of Laws (KCL & KVL):

KCL is used for node analysis of any given circuit where the equations are developed to solve for the node voltages and hence, thereafter the branch currents.

KVL is used for mesh analysis of any given circuit where the equations are developed to solve for mesh currents and hence, thereafter the voltage drops across branches of the circuit from which the node voltages can be arrived.

Alternatively, once can either use KVL or KCL for analysis of any given circuit (both DC & AC).

In Fig.1, it's seen that the DC Circuit consists of two meshes. If we choose to solve the circuit by Mesh Analysis using KVL, we can assume two Mesh Currents as I₁

& I_2, as shown in Fig.2 below:

                                                                         Fig.2

Circuit Analysis:

The two mesh currents - I₁& I_2, as shown here are assumed in clock-wise directions. Now, there are chances that, any one of the current may flow in opposite direction in the actual case of the circuit, since we have just assumed the directions of the current. In such cases, we just get negative values of the current, after solving it, as shown below in Fig.3

                                                                         Fig.3
Here as shown above, as said earlier, the value for the current I₂ solved is -0.5 A, after writing down the mesh equations.This just means that the current I₂ is flowing in counter-clock-wise direction in actual case of the circuit.

Developing Mesh Equations Guide (Assigning Signs): 

Let's write the equation for the mesh current I_1:
For the assumed clock-wise direction of I₁, the current sees the DC voltage as -10 V, because the current first passes through the minus side of the DC voltage source. That's why, at the start of this chapter, I'd said it's ±10V. Hence, while writing the mesh equation for current I₁, the value 10 is assigned minus sign.
While the current I₁, flows through the mesh, it develops drops across the resistors as "+2I₁+2I₁+2I₁". Now, one of the branches, of 2Ω resistor, is common to both Mesh 1 & 2. Here the other mesh current I₂, also flows through the 2Ω resistor branch, but from bottom to top,i.e. in the opposite direction of  I_1. Hence it's assigned minus sign,as shown in Fig. 4 below:

                                                                             Fig.4

Hence the, mesh equation for Mesh 1 is:

-10+2I1+2I1-2I2+2I1=0

=> 3I1-I2=5 - (1)

Similarly, for the second mesh, the mesh equation is:

2I2-2I1+2I2+5 = 0

=> 2I1-4I2=5- (2)

solving (1) & (2), 

I1= 1.5 A

I2= -0.5 A       - Here the current is negative, since in actual case of the given circuit the current flows in the counter-clock-wise direction.

SOLUTION (With Actual Flow of Currents):

  Here, what you are looking at, is the actual flow of current in Fig.5, as below:

                                                                              Fig.5

In the above illustration, what we had seen, is application of KVL to solve for mesh equations and the mesh currents. The branch current of 2A (1.5+0.5) is illustrated using KCL as below in Fig.6

                                                                              Fig.6

Here, we can see that the branch current is just the algebraic summation of the mesh currents I1&I2, which is illustrated by a small demonstration of KCL at Node N of the Dc Circuit.

For those who need an actual video demo of solving the above DC circuit, without missing any points, please do view below link:

For a more complex DC Circuit, with 3 Meshes, I've illustrated with "Node Voltage Analysis" and an easy method to solve large number of simultaneous equations with 7 unknown variables, as in below given link.