Kirchhoff's Laws of Electrical Network

Terms defined before the laws

• Junction: any point in an electric circuit where two or more conductors are joined together.
• Loop (mesh): any closed conducting path in an electric network.
• Branch: any part of the network that lies between two junctions.
• In Fig. 9.1 there are two junctions, labelled a and b, and three branches — the three possible paths 1, 2 and 3 from a to b.
• The two laws below apply to a steady current flowing through an electrical network of resistors.

9.2.1 Kirchhoff's first law (current law, junction law)

Statement

• The algebraic sum of the currents at a junction in an electrical network is zero.

$$\sum_{i=1}^{n} I_i = 0$$

• where $I_i$ is the current in the $i^{th}$ conductor at a junction having $n$ conductors.

Sign convention

• The currents arriving at the junction are considered positive.
• The currents leaving the junction are considered negative.

The diagram in words (Fig. 9.2)

• A single point P at which six conductors meet. Three of the arrows ($I_1$, $I_3$, $I_4$) point inwards, towards P; three ($I_2$, $I_5$, $I_6$) point outwards, away from P.

Applying it

$$I_1 - I_2 + I_3 + I_4 - I_5 - I_6 = 0$$

• Arriving currents $I_1$, $I_3$, $I_4$ are positive; leaving currents $I_2$, $I_5$, $I_6$ are negative. Rearranged,

$$I_1 + I_3 + I_4 = I_2 + I_5 + I_6$$

• so the total current flowing towards the junction equals the total current flowing away from the junction.

Why it is true

• The junction law is consistent with the conservation of electric charge. Some charge is received per unit time due to the currents arriving at the junction; for charge to be conserved, the same amount of charge must leave the junction per unit time, since charge cannot accumulate at a point.

9.2.2 Kirchhoff's second law (voltage law, loop theorem)

Statement

• The algebraic sum of the potential differences (products of current and resistance) and the electromotive forces (emfs) in a closed loop is zero.

$$\sum IR + \sum E = 0$$

Sign convention

• While tracing a loop through a resistor, if we travel along the direction of the conventional current, the potential difference across that resistance is taken as negative. If the loop is traced against the direction of the conventional current, the potential difference across that resistor is taken as positive.
• The emf of an electrical source is positive while tracing the loop within the source from its negative terminal to its positive terminal. It is taken as negative while tracing within the source from the positive terminal to the negative terminal.

The diagram in words (Fig. 9.3)

• A two-loop network. The left loop ABFGA contains the source $E_1$ and resistances $R_1$, $R_3$ carrying $I_1$, with $R_5$ in the central branch carrying $I_3$. The right loop BFDCB contains the source $E_2$ and resistances $R_2$, $R_4$ carrying $I_2$, sharing the same central branch $R_5$.

Applying it

• Loop ABFGA traced clockwise:

$$-I_1R_1 - I_3R_5 - I_1R_3 + E_1 = 0 \qquad \therefore\ E_1 = I_1R_1 + I_3R_5 + I_1R_3$$

• Loop BFDCB traced anticlockwise:

$$-I_2R_2 - I_3R_5 - I_2R_4 + E_2 = 0 \qquad \therefore\ E_2 = I_2R_2 + I_3R_5 + I_2R_4$$

Why it is true

• The voltage law is consistent with the law of conservation of energy. The algebraic sum of the emfs is the electrical energy supplied per unit charge by the sources; by conservation of energy this energy must reappear as potential differences across the electrical elements in the loop.

Steps usually followed while solving a problem using Kirchhoff's laws

• Choose some direction for the currents.
• Reduce the number of variables using Kirchhoff's first law.
• Determine the number of independent loops.
• Apply the voltage law to all the independent loops.
• Solve the equations obtained simultaneously.
• If the answer for a current variable is negative, the conventional current flows in the direction opposite to the one chosen.

Worked pattern: two cells in parallel across a resistance

• Standard question: two batteries of emfs $E_1$, $E_2$ with internal resistances $r_1$, $r_2$ are connected in parallel across an external resistance $R$, sending current in the same direction through $R$.
• Let $I_1$ and $I_2$ be the currents through the two cells; the current through $R$ is $I_1+I_2$.
• Loop through cell 1 and $R$: $E_1 = I_1r_1 + (I_1+I_2)R$.
• Loop through cell 2 and $R$: $E_2 = I_2r_2 + (I_1+I_2)R$.
• Solve the pair simultaneously; the potential difference across $R$ is $(I_1+I_2)R$.
• For $E_1 = 4$ V, $r_1 = 1\ \Omega$, $E_2 = 2$ V, $r_2 = 2\ \Omega$, $R = 5\ \Omega$, this gives a current through the external resistance of $\dfrac{10}{17}$ A. A negative value for one of $I_1$, $I_2$ simply means that cell is being driven backwards by the other.

Standard results collected

• $\sum_{i=1}^{n}I_i = 0$ at a junction; charge conservation.
• $\sum IR + \sum E = 0$ round a loop; energy conservation.
• $I = \dfrac{E}{R+r}$ and $V = E - Ir$ for a single cell driving a single resistance; a voltmeter of resistance $R_V$ across such a cell reads $V = \dfrac{ER_V}{R_V+r}$, always less than $E$.
• Kirchhoff's laws are applicable to both DC and AC circuits; for AC, instantaneous values of currents and voltages are used.