KIRCHHOFF'S LAWS AND SIMPLE APPLICATIONS :
Question : State and explain Kirchhoff's rules with the help of simple applications.
Kirchhoff's laws are used to find the currents and voltages in different parts of the circuit.
Kirchhoff's First Law or Rule (The Junction Law or Kirchhoff's Current Law) :
It states that the sum of all the currents entering any point (or junction) must be equal to the sum of all currents leaving that point (junction).
The algebraic sum of all the currents meeting at a point (or junction) in a closed electrical circuit is zero.
That is,
$ \Sigma I = 0 $
Consider a point or junction O in an electrical circuit (Figure 37). Let $I_1$, $I_3$ be the currents entering the point O and $I_2$, $I_4$, $I_5$ be the currents leaving the point O. Then according to Kirchhoff's first law or junction law,
$I_1 + I_3 = I_2 + I_4 + I_5 \quad \dots(i)$
or
$ I_1 + I_3 + (-I_2) + (-I_4) + (-I_5) = 0 $
or
$ I_1 + I_3 - I_2 - I_4 - I_5 = 0$
or
$\Sigma I = 0 \dots(ii) $
The currents entering the point can be taken as positive while currents leaving the point can be taken as negative.
Note: KIRCHHOFF'S FIRST LAW IS BASED ON THE LAW OF CONSERVATION OF CHARGES i.e. on the fact that charges do not remain accumulated at a junction of a circuit.
Kirchhoff's Second Law (The Loop Law or Kirchhoff's Voltage Law) :
It states that the algebraic sum of all voltages i.e., the potential differences across all elements and e.m.fs of all sources in any closed electrical circuit is zero.
That is,
$ \Sigma E + \Sigma \Delta V = 0 $
Sign Convention:
1. In a given electric circuit, when a resistor R is travelled (traversed) in the direction of the current I, the potential difference $\Delta V$ across the resistor may be taken as negative and when a resistor is travelled (traversed) in the direction opposite to the direction of current I, potential difference $\Delta V$ across the resistor may be taken as positive.
2. In a given electric circuit, when a source of e.m.f. say a cell is travelled (traversed) from - to + terminal, the e.m.f. of the source may be taken as positive and when a source of e.m.f. is travelled (traversed) from + to - terminal, the e.m.f. the source may be taken as negative.
Examples :
Let us calculate the currents in the two resistances :
Applying Kirchhoff's loop law to the closed loop ABCDA.
$$ -2i_1 - 1.5 - 3i_2 + 3 = 0 \quad \text{or} \quad 2i_1 + 3i_2 = 1.5 \quad \dots(i) $$
Current $i_1$ and $i_2$ are due to the net effect of two sources of e.m.f. and resistances 2 $\Omega$ and 3 $\Omega$ are in series, so
$ i_1 = i_2 = i $
Using equation (i), eqn. (i) becomes
$ 2i + 3i = 1.5 $
or
$ 5i = 1.5 $
or
$ i = \frac{1.5}{5} $
Or
$ i = 0.3 \text{ A} $
or
$ i = 0.3 \text{ A} $
Hence,
$ i_1 = 0.3 \text{ A} $
$ i_2 = 0.3 \text{ A} $
Note : KIRCHHOFF'S Second LAW IS BASED ON THE LAW OF CONSERVATION OF Energy.