THEORY

Kirchoff's voltage law states that the algebraic sum of all branch voltages around any closed path in a circuit is always zero at all instants of time. When the current passes through a resistor there is a loss of energy and therefore, a voltage drop. In any element, the current always flows from higher to lower potential. Consider the circuit in figure 1. It is customary to take the direction of current I as indicated in figure. That is it leaves the positive terminal of voltage source and enters into the negative terminal.

As the current passes through the circuit ,the sum of the voltage drop around the loop is equal to the total voltage in that loop. Here the polarity are attributed to the resistors to indicate that the voltages at points a ,c and e are more than the voltages at b ,d and f, respectively, as current passes from a to f.

Therefore VS=V1+V2+V3

Consider the problem of finding out the current supplied by source V in the circuit shown in figure 2. Our first step is to assume the reference current direction and to indicate the polarities for different elements (see figure 3).

By using Ohm's law we find the voltage across each resistor as follows

VR1=IR1,VR2=IR2,VR3=IR3¬

Where VR1,VR2 and VR3 are the voltages across R1 R2 and R3 respectively. Finally by applying Kirchoff's law we can form the equation

V=VR1+ VR2+ VR3

V=IR1 +IR2 +IR3¬

From the above equation the current delivered by source is given by

I = V/(R1+R2+R3)