Kirchhoff's laws
Section: D.C. Circuits | Syllabus: Cambridge AS Level Physics (9702)
Kirchhoff's First Law (Junction Rule)
Kirchhoff's First Law The algebraic sum of the currents at a junction is zero.Equivalently: The sum of currents entering a junction equals the sum of currents leaving the junction.
This law is a consequence of the conservation of charge. Charge cannot be created or destroyed, so charge carriers cannot accumulate at a junction. All charge entering must leave.
Mathematical Statement Σ I_in = Σ I_outOr using sign convention (positive for entering, negative for leaving):
Σ I = 0 FIG 10.6: Current at a Junction
Show a junction point with five wires meeting. Label currents I₁ and I₂ with arrows pointing INTO the junction, and I₃, I₄, I₅ with arrows pointing OUT of the junction. Include the equation: I₁ + I₂ = I₃ + I₄ + I₅
Worked Example: Junction CurrentsProblem: At a junction, currents of 500 mA, 700 mA, and 400 mA flow in. A current of 200 mA flows out. What is the magnitude and direction of the fifth current?
Solution:
Using Σ I_in = Σ I_out: 500 + 700 + 400 = 200 + I_5 1600 = 200 + I_5 I_5 = 1400 mA = 1.4 A (leaving the junction)Analogy: River Splitting Around an Island
Consider a river approaching an island. If the river current (rate of water flow) is 300 m³s⁻¹ before the split, and 100 m³s⁻¹ flows down one channel, then 200 m³s⁻¹ must flow down the other. When the channels recombine, the total is again 300 m³s⁻¹. Water is conserved - just as charge is conserved in an electrical circuit.
Kirchhoff's Second Law (Loop Rule)
Kirchhoff's Second Law The algebraic sum of the potential differences around any closed loop in a circuit is zero.This law is a consequence of the conservation of energy. As charge moves around a complete loop, the energy gained (from sources of e.m.f.) must equal the energy transferred (in components like resistors).
Mathematical Statement Σ = Σ IROr: Σ V = 0 around any closed loop
Sign Conventions for Loop Analysis
- Choose a direction to trace around the loop (clockwise or anticlockwise)
- E.m.f.: Positive if you pass from − to + through the source; negative if + to −
- Resistor p.d.: Positive if you traverse in the opposite direction to current flow; negative if in the same direction as current
Show a simple circuit loop containing: a battery (e.m.f. = ε, internal resistance = r) connected to two resistors R₁ and R₂ in series. Draw a clockwise arrow indicating the loop direction. Label the current I flowing around the loop. Show that: ε = Ir + IR₁ + IR₂
Energy Perspective
Consider 1 coulomb of charge moving around a complete circuit:
- The charge gains energy equal to the e.m.f. (in joules) as it passes through the power source
- The charge loses this energy as it passes through resistors (converted to heat, light, etc.)
- When returning to the starting point, net energy change = 0
Analogy: Ball on a Hill
Consider a ball being pushed uphill from point A to point B, gaining gravitational potential energy mgΔ h. As the ball rolls downhill from B through C to D (at the same height as A), it loses this energy in stages:
- Energy lost from B to C: mgΔ h_1
- Energy lost from C to D: mgΔ h_2
- Total energy lost: mgΔ h_1 + mgΔ h_2 = mgΔ h (equals energy gained)
Show a hill profile with points A (bottom left), B (top), C (partway down), D (bottom right, same level as A). Show ball being pushed up from A to B (gaining energy = mg∆h), then rolling down through C to D (losing energy mg∆h₁ then mg∆h₂). Label the heights ∆h, ∆h₁, and ∆h₂.
This is exactly analogous to charge in a circuit:
- Charge gains electrical potential energy passing through the battery (like being pushed uphill)
- Charge loses this energy passing through resistors (like rolling downhill)
- Total energy gained = total energy lost (conservation of energy)
Imagine two identical balls rolling down opposite sides of the same hill - they both lose the same amount of gravitational potential energy. Similarly, charge passing through parallel branches loses the same electrical energy (same p.d.) regardless of which path it takes.
Reversed Cells
If a cell is reversed so that it has opposite polarity to other cells in a series battery, its p.d. has a negative value and must be subtracted from the total e.m.f.
Derivation: Resistors in Series
Using Kirchhoff's laws, we can derive the formula for combined resistance of resistors in series.
FIG 10.9: Three Resistors in SeriesShow a circuit with a power source of p.d. V connected to three resistors R₁, R₂, R₃ in series. Label the current I through all components. Label the p.d. across each resistor as V₁, V₂, V₃ respectively.
Derivation
From Kirchhoff's first law: Since there are no junctions, the same current I flows through all resistors.
From Kirchhoff's second law: The total p.d. equals the sum of p.d.s across each resistor:
V_T = V_1 + V_2 + V_3Applying V = IR to each resistor:
V_1 = IR_1, V_2 = IR_2, V_3 = IR_3Substituting:
V_T = IR_1 + IR_2 + IR_3 = I(R_1 + R_2 + R_3)If R_T is the combined resistance such that V_T = IR_T, then:
Resistors in Series R_T = R_1 + R_2 + R_3 + ... Physical Insight - Connection to ResistivityAdding resistors in series is like increasing the length of a wire. Recall that R = (ρ l/A), where ρ is resistivity, l is length, and A is cross-sectional area. Doubling the length doubles the resistance. A resistor is simply an ohmic conductor (like a wire), so adding resistors in series increases the total "length" the charge must travel through.
Derivation: Resistors in Parallel
Using Kirchhoff's laws, we can derive the formula for combined resistance of resistors in parallel.
FIG 10.10: Three Resistors in ParallelShow a circuit with a power source of p.d. V connected to three resistors R₁, R₂, R₃ in parallel. Show the total current I entering a junction and splitting into I₁, I₂, I₃ through each resistor. Indicate that all resistors have the same p.d. V across them.
Derivation
From Kirchhoff's second law: Tracing a loop through any two parallel resistors, the p.d. across each must be equal (same two points):
V = V_1 = V_2 = V_3From Kirchhoff's first law: The total current divides at the junction:
I_T = I_1 + I_2 + I_3Applying I = V/R to each branch:
I_1 = (V/R_1), I_2 = (V/R_2), I_3 = (V/R_3)Substituting:
I_T = (V/R_1) + (V/R_2) + (V/R_3) = V((1/R_1) + (1/R_2) + (1/R_3))If R_T is the combined resistance such that I_T = V/R_T, then:
Resistors in Parallel (1/R_T) = (1/R_1) + (1/R_2) + (1/R_3) + ... Physical Insight - Connection to ResistivityAdding resistors in parallel is like increasing the cross-sectional area of a wire. From R = (ρ l/A), doubling the cross-sectional area halves the resistance. When resistors are in parallel, charge has more paths to flow through - effectively a larger "cross-sectional area" for current flow, reducing total resistance.
ImportantAlways remember to take the reciprocal at the end to find R_T, not 1/R_T!
Worked Example: Lamp in Parallel with ResistorProblem: A lamp rated at 12 V, 48 W is connected in parallel with a 3 Ω resistor. Find their combined resistance.
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