Kirchhoff's laws

1. Overview

Kirchhoff’s laws are the fundamental principles governing the behavior of electric circuits, derived directly from the universal laws of conservation of charge and conservation of energy. Kirchhoff’s First Law (the Junction Rule) ensures that charge is neither created nor destroyed at any point in a circuit. Kirchhoff’s Second Law (the Loop Rule) ensures that the total energy supplied to a circuit by electromotive forces (e.m.f.s) is exactly accounted for by the energy transferred in the components (potential differences). Together, these laws provide the mathematical basis for deriving the formulas for combined resistance and for solving complex networks that cannot be simplified by basic series or parallel rules alone.

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Key Definitions

• Kirchhoff’s First Law: The sum of the currents entering any junction in a circuit is equal to the sum of the currents leaving that junction.
• Kirchhoff’s Second Law: In any closed loop of a circuit, the algebraic sum of the e.m.f.s is equal to the algebraic sum of the potential differences (p.d.s).
• Conservation of Charge: A fundamental law of physics stating that the total electric charge in an isolated system remains constant. Charge cannot be created or destroyed, only transferred.
• Conservation of Energy: A fundamental law stating that energy cannot be created or destroyed, only transformed from one form to another. In a circuit, chemical energy in a source is transformed into electrical energy, which is then transformed into other forms (heat, light) in the load.
• Junction: A point in a circuit where three or more circuit paths (conductors) meet.
• Closed Loop: Any complete conducting path that starts and ends at the same point in a circuit.
• Electromotive Force (e.m.f.): The energy transferred by a source to each unit of charge that passes through it.
• Potential Difference (p.d.): The energy transferred from each unit of charge to the components in a circuit.

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Content

3.1 Kirchhoff’s First Law (The Junction Rule)

Kirchhoff’s First Law is a statement of the conservation of charge.

Mathematical Expression: $$\sum I_{\text{in}} = \sum I_{\text{out}}$$

Physical Basis: Current is defined as the rate of flow of charge ($I = \Delta Q / \Delta t$). In a steady-state circuit, charge does not accumulate at any point or junction. Therefore, the total charge flowing into a junction per second must equal the total charge flowing out of that junction per second. If this were not true, the junction would become increasingly charged over time, which does not happen in stable DC circuits.

Application: When analyzing a junction, assign a positive sign to currents entering and a negative sign to currents leaving (or vice versa, as long as you are consistent).

• Example: If $I_1$ and $I_2$ enter a junction, and $I_3$ and $I_4$ leave it: $$I_1 + I_2 = I_3 + I_4$$ or $$I_1 + I_2 - I_3 - I_4 = 0$$

3.2 Kirchhoff’s Second Law (The Loop Rule)

Kirchhoff’s Second Law is a statement of the conservation of energy.

Mathematical Expression: $$\sum \mathcal{E} = \sum V$$ or $$\sum \mathcal{E} = \sum (IR)$$

Physical Basis: Potential difference and e.m.f. are defined as work done per unit charge ($V = W/Q$). As a unit of charge moves around a complete closed loop, it gains energy from the sources of e.m.f. and loses energy as it passes through resistors or other components. Because energy is conserved, the total energy gained by the charge must equal the total energy dissipated by the charge when it returns to its starting point. Thus, the net change in potential energy around any closed loop is zero.

Sign Conventions for Loop Analysis: To apply $\sum \mathcal{E} = \sum (IR)$ correctly, you must follow a consistent sign convention:

1. Choose a loop direction: Either clockwise or anticlockwise.
2. For e.m.f.s ($\mathcal{E}$):
   • If you move through a battery from the negative terminal to the positive terminal, the e.m.f. is positive ($+\mathcal{E}$).
   • If you move from positive to negative, the e.m.f. is negative ($-\mathcal{E}$).
3. For p.d.s ($IR$ terms):
   • If you move through a resistor in the same direction as the assigned current, the $IR$ term is positive (it is a "drop" in potential).
   • If you move against the direction of the assigned current, the $IR$ term is negative.

3.3 Derivation: Combined Resistance in Series

Objective: To derive the formula $R = R_1 + R_2 + \dots$ using Kirchhoff’s Laws.

1. Circuit Setup: Consider three resistors $R_1, R_2,$ and $R_3$ connected in series to a source of e.m.f. $V$.
2. Apply Kirchhoff’s First Law: In a series circuit, there are no junctions. Therefore, the current $I$ must be the same through every resistor. $$I_{total} = I_1 = I_2 = I_3 = I$$
3. Apply Kirchhoff’s Second Law: For the single closed loop, the sum of e.m.f.s equals the sum of p.d.s across the resistors. $$V = V_1 + V_2 + V_3$$
4. Substitute Ohm’s Law: Since $V = IR$ for each component: $$V = IR_1 + IR_2 + IR_3$$
5. Define Combined Resistance: Let $R$ be the total (equivalent) resistance such that $V = IR$. $$IR = IR_1 + IR_2 + IR_3$$
6. Simplify: Divide the entire equation by the common current $I$: $$R = R_1 + R_2 + R_3$$

3.4 Derivation: Combined Resistance in Parallel

Objective: To derive the formula $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$ using Kirchhoff’s Laws.

1. Circuit Setup: Consider three resistors $R_1, R_2,$ and $R_3$ connected in parallel across a source of e.m.f. $V$.
2. Apply Kirchhoff’s Second Law: Consider loops containing the source and only one resistor at a time. For each loop, the e.m.f. must equal the p.d. across that branch. Therefore, the potential difference $V$ is the same across each resistor. $$V_{total} = V_1 = V_2 = V_3 = V$$
3. Apply Kirchhoff’s First Law: At the junction where the main circuit splits, the total current $I$ is the sum of the currents in the individual branches. $$I = I_1 + I_2 + I_3$$
4. Substitute Ohm’s Law: Since $I = V/R$ for each component: $$\frac{V}{R} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$$
5. Simplify: Divide the entire equation by the common potential difference $V$: $$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$

3.5 Worked Example 1 — Complex Resistor Network

Question: A $24.0\text{ V}$ DC supply is connected to a $10.0\text{ }\Omega$ resistor ($R_1$) in series with a parallel combination of two resistors: $R_2 = 30.0\text{ }\Omega$ and $R_3 = 60.0\text{ }\Omega$. Calculate the current flowing through the $30.0\text{ }\Omega$ resistor.

Step 1: Find the equivalent resistance of the parallel section ($R_p$) $$\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3}$$ $$\frac{1}{R_p} = \frac{1}{30.0} + \frac{1}{60.0} = \frac{2}{60.0} + \frac{1}{60.0} = \frac{3}{60.0}$$ $$R_p = \frac{60.0}{3} = 20.0\text{ }\Omega$$

Step 2: Find the total circuit resistance ($R_{total}$) $$R_{total} = R_1 + R_p$$ $$R_{total} = 10.0 + 20.0 = 30.0\text{ }\Omega$$

Step 3: Find the total current ($I_{total}$) leaving the supply $$I_{total} = \frac{V}{R_{total}} = \frac{24.0}{30.0} = 0.80\text{ A}$$

Step 4: Find the p.d. across the parallel section ($V_p$) Using Kirchhoff's Second Law for the loop containing $R_1$ and $R_p$: $$V_p = I_{total} \times R_p = 0.80 \times 20.0 = 16.0\text{ V}$$ (Alternatively: $V_p = 24.0 - (0.80 \times 10.0) = 16.0\text{ V}$)

Step 5: Find the current through $R_2$ ($I_2$) $$I_2 = \frac{V_p}{R_2} = \frac{16.0}{30.0} = 0.533\text{ A}$$

Answer: $0.53\text{ A}$ (to 2 s.f.)

3.6 Worked Example 2 — Applying the Loop Rule with Two Sources

Question: A single loop circuit contains two batteries and one resistor. Battery A has an e.m.f. of $12.0\text{ V}$. Battery B has an e.m.f. of $4.0\text{ V}$. The batteries are connected "positive terminal to positive terminal". The total resistance in the loop is $5.0\text{ }\Omega$. Calculate the current in the circuit and state its direction.

Step 1: Identify the net e.m.f. ($\sum \mathcal{E}$) Since the batteries are opposing each other (positive to positive), their e.m.f.s subtract. Assume the direction of the stronger battery (Battery A) is the positive loop direction. $$\sum \mathcal{E} = \mathcal{E}_A - \mathcal{E}_B = 12.0 - 4.0 = 8.0\text{ V}$$

Step 2: Apply Kirchhoff’s Second Law $$\sum \mathcal{E} = \sum (IR)$$ $$8.0 = I \times 5.0$$

Step 3: Solve for $I$ $$I = \frac{8.0}{5.0} = 1.6\text{ A}$$

Step 4: Determine direction The current flows out of the positive terminal of the $12.0\text{ V}$ battery.

Answer: $1.6\text{ A}$ in the direction of the $12.0\text{ V}$ battery's e.m.f.

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Key Equations

Law / Formula	Equation	Data Sheet?
Kirchhoff's 1st Law	$\sum I_{\text{in}} = \sum I_{\text{out}}$	No
Kirchhoff's 2nd Law	$\sum \mathcal{E} = \sum V$	No
Series Resistance	$R = R_1 + R_2 + \dots$	Yes
Parallel Resistance	$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$	Yes
Ohm's Law	$V = IR$	No

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Common Mistakes to Avoid

• ❌ Wrong: Stating that Kirchhoff’s First Law is about energy conservation.
  • ✓ Right: Kirchhoff’s First Law is about charge conservation; the Second Law is about energy conservation.
• ❌ Wrong: Forgetting to calculate the reciprocal at the end of a parallel resistance calculation (e.g., leaving the answer as $1/R = 0.2$).
  • ✓ Right: Always perform the final step: $R = 1 / 0.2 = 5.0\text{ }\Omega$.
• ❌ Wrong: Adding e.m.f.s of batteries regardless of their orientation.
  • ✓ Right: Check the polarity. If batteries face each other (+ to +), subtract the e.m.f.s. If they are in series (+ to -), add them.
• ❌ Wrong: Assuming current is the same in all branches of a parallel circuit.
  • ✓ Right: Current splits at junctions. Only the potential difference is the same across parallel branches.
• ❌ Wrong: Using the wrong units (e.g., $k\Omega$ instead of $\Omega$).
  • ✓ Right: Convert all prefixes to base SI units ($1.2\text{ k}\Omega = 1200\text{ }\Omega$) before calculating to avoid power-of-ten errors.

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Exam Tips

1. The "Starting Principle": When asked to derive the resistance formulas, you must state the Kirchhoff Law you are using as your first step. For series, start with $\sum \mathcal{E} = \sum V$. For parallel, start with $\sum I_{\text{in}} = \sum I_{\text{out}}$.
2. Draw the Loop: In complex problems, physically draw a circular arrow in the center of the loop to indicate your chosen direction. This helps you stay consistent with signs for $IR$ drops and e.m.f.s.
3. Internal Resistance: Remember that Kirchhoff’s Second Law applies to the entire loop, including the internal resistance of the battery. Treat internal resistance ($r$) just like any other resistor ($Ir$ drop).
4. Significant Figures: Cambridge 9702 typically requires answers to 2 or 3 significant figures. Look at the data provided in the question; if the e.m.f. is $12.0\text{ V}$ (3 s.f.), your answer should likely be 3 s.f.
5. Keywords in Definitions: When defining Kirchhoff’s First Law, always use the word junction. For the Second Law, always use the phrase closed loop. These are often required for the first mark in a multi-part question.
6. Verification: If time permits, check your junction currents. The sum of currents entering must exactly equal the sum leaving. If they don't, you have a sign error in your loop equations.