Kirchhoff’s Laws

KIRCHHOFF'S LAWS

Junction Rule:

It is based on the law of conservation of charge. At a junction in a circuit the sum of incoming currents is always equal to the sum of outgoing currents. In otherwords the algebraic sum of the currents at a junction is zero.

Loop rule

The algebraic sum of the changes in potential around any closed path is zero. It is based on the law of conservation of energy.

$\bullet$In case of a resistor of resistance 'R' potential will decrease in the direction of current. Hence, for the shown conductor

V_a – V_b = IR

$\bullet$ For an emf source, the potential changes will be obtained as illustrated below,

Emf = $\epsilon$, internal resistance = r

V_a – V_b = $\epsilon$ + ir

Emf = $\epsilon$, internal resistance =

V_a – V_b = -$\epsilon$ + ir

Students can use any sign convention which they find easy.

Example 1: In the series circuit shown, E,F,G,H are cells of emf 2V,1V,3V and 1V respectively, and their internal resistance are 2, 1, 3 and 1$\Omega$ respectively.

Calculate

(i) the potential difference between B and D and

(ii) the potential difference across the terminals of each of the cells G and H.

Solution: Let us redraw the circuit.

At junction D, we have applied the junction rule, whereby we get current in DB as shown.

Loop BADB

2I₁ - 2 + 1 + I₁ + 2 (I₁ - I₂) = 0

$\Rightarrow$ 5I₁ - 2I₂ = 1

Loop DCBD

-3+3I₂+I₂+1-2(I₁-I₂)=0 $\Rightarrow$ 6I₂ – 2I₁ = 2

$\Rightarrow I_1=\frac{5}{13}A,I_2=\frac{6}{13}A$

(i) V_BD = 2(I₁) - 2 + 1 + I₁= 3 I₁ - 1 = $3\left(\frac{5}{13}\right)-1=\frac{2}{13}V$

(ii) Terminal voltage of G = |-3 + 3.I₂| = $\left|-3+3\left(\frac{6}{13}\right)\right|$ = $\frac{21}{13}V$

Terminal voltage of H = $\left|\frac{6}{13}(I)+1\right|=\frac{19}{13}V$.

GROUPING OF RESISTANCES

Resistance in series

Let the equivalent resistance between A & B equals R_eq , by definition.

R_eq = $\frac{V}{I}$ . . . (1)

Using Kirchoff's 2nd rule for the loop shown in figure,

V = IR₁ + IR₂ + IR₃ . . . (2)

From (1) and (2) R_eq = R₁ + R₂ + R₃

Resistance in parallel

Here again, R_eq = $\frac{V}{I}$ . . . (1)

I = i₁ + i₂ + i₃ = $\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}$ . . . (2)

From (1) and (2)

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Example 2: Find the equivalent resistance between A and B in the circuit shown here. Every resistance shown here has a magnitude of 2 $\Omega$.

Solution: Points C, O & D are at the same potential. Therefore, resistances AO, AC and AD are in parallel . Similarly BC, BO and BD are in parallel.

Similarly BC, BO and BD are in parallel.

$\therefore$ R_AB = $\frac{1}{3}$x(2$\Omega$) + $\frac{1}{3}$ x (2$\Omega$)

= $\frac{4}{3}$$\Omega$ = 1.33 $\Omega$.