Kirchoff’s Point Law or Current Law (KCL)

It can also be stated in another
way which is, the total current leaving a junction is equal to the total
current entering that junction. This is obviously true because there is no
accumulation or generation of charges at the junction of the network.

Figure 1a shows a number of conductors
carrying currents meeting at a point A. Some conductors have currents
approaching point A while others have currents leaving the point. If we assume
all incoming currents to the point to be positive and the outgoing currents to
be negative, we have;

I1 + (-I2)
+ (-I3) + I4 + (-I5) = 0

Or I1 – I2 –
I3 + I4 – I5 = 0

Or I1 + I4  = I2 + I3 + I5

Or Incoming currents = Outgoing
currents

Similarly, in figure 1b, for the
point A,

I + (-I1) + (-I2)
+ (-I3) + (-I4) = 0

Or I = I1 + I2
+ I3 + I4

Using the above expressions,
Kirchoff’s Current Law can be stated in a general term as

∑ I = 0  (at a junction)

Example 1

Using Kirchoff’s Current Law and
Ohm’s Law, find the magnitude and polarity of voltage V in figure 2 above.

Solution

Let us arbitrarily choose the directions of currents Ia,
Ib and Ic and polarity of V as shown in figure 3 below.

Applying KCL to node A, we have

Ia + 25 + Ib – Ic – 10 = 0

Or Ia + Ib – Ic = -15     – – – – – – – – – – (1)

Applying Ohm’s law to the three resistive branches, we have

Ia = V/5, 
Ib = V/8,  Ic
= – V/6

Please note that the negative
sign in Ic indicates the fall in potential considering the polarity
of the voltage and the direction of current through Ic.

Substituting the above values of Ia,
Ib and Ic into (1) above we have 

V/5 + V/8 – (-V/6) = -15

V/5 + V/8 + V/6 = -15

                  V = -30.5V

The magnitude of V is 30.5V and
the polarity is from B to A and not from A to B as shown in figure 3 above.