Kirchhoff's second law
Key idea: Potential rises and drops sum to zero around a closed circuit loop. Use one sign convention and include each source's e.m.f. and internal resistance.
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Guided steps
- Relate each potential change to energy transferred per coulomb.
- Choose the traversal direction separately from each current reference arrow.
- Decide the resistor term when following conventional current.
- Write both traversal equations and show that they predict the same current.
- Calculate the loop current before comparing the full method.
- Write the loop balance and verify both resistor voltage drops.
- Include internal resistance in the series loop before calculating current.
- Calculate the net source rise, current magnitude and direction.
- Follow the coupled junction and loop equations, keeping the shared current distinct.
- Solve all three signed currents after replacing one source.
- Compare the complete equation solution and verify both voltage balances.
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Around any closed loop in a circuit, the rises in potential must balance the drops. Kirchhoff’s second law turns this into one equation for each loop. Together with the first law, it lets you find the currents in a circuit that is not a simple series or parallel combination.
Kirchhoff’s laws are not explicitly listed in the H2 syllabus, but they are a common enrichment tool for solving non-trivial D.C. circuit networks.
The loop rule
Kirchhoff’s second law states that around any closed loop, the algebraic sum of the potential changes is zero:
It is a statement of conservation of energy. Potential difference is energy transferred per unit charge. A charge that goes round a loop returns to the same point, so its net change in electrical potential energy is zero. That is why e.m.f. rises and resistive drops can be added in one equation.
Signs around a loop
Choose a direction to go round the loop, then use one sign convention throughout:
- across a source from − to +: + ε
- across a source from + to −: -ε
- across a resistor in the direction of the conventional current: -IR
- across a resistor against the current: + IR
Spot the mistake 1
The sign of a resistor term
Learner claim
Try this before viewing the solution
Show solution step by step
Identify the physical change
Method
Potential decreases through a passive resistor in the current direction.Reason
Electrical energy per unit charge is transferred to thermal energy.Working
Answer
Δ V = V after -V before < 0Write the loop term
Method
The term is -IR.Reason
IR is the magnitude of the resistive potential drop.Working
Answer
Δ VR = -IR
Common mistake
Check your understanding 1 Reversing loop direction
Two opposing sources of 10 V and 4 V drive current through 3.0 Ω total resistance. Define I in the direction driven by the larger source. Write a loop equation for each traversal direction and calculate I from both.
Show answer
Traversing with the defined current gives 10-4-3.0I = 0, so I = 2.0 A. Traversing the other way gives -10 + 4 + 3.0I = 0, again giving I = 2.0 A in the defined direction. Reversing the traversal reverses every term; it does not reverse the physical current or its reference arrow.
Single loops
In a single loop, the same current passes through every component. The loop equation then sets the sum of the source e.m.f.s equal to the sum of the IR drops, including any internal resistance.
Exam-style question 1
Two resistors in series
Examination question
An ideal 12 V source is connected in series with R₁ = 4.0 Ω and R₂ = 8.0 Ω. Use Kirchhoff’s second law to find the current. [3 marks]
Try this before viewing the solution
Show the steps Optional working
Show solution step by step
Write the loop equation
1 markMethod
12-IR₁-IR₂ = 0.Reason
The source rise balances both series-resistor drops.
Working
Answer
12-I(4.0)-I(8.0) = 0Collect the drops
1 markMethod
12 = 12I.Reason
The same current passes through both series resistors.
Working
Answer
I(4.0 + 8.0) = 12Calculate
1 markMethod
I = 1.0 A.Reason
A 12 V rise is balanced by the drop across 12 Ω total resistance.
Working
Answer
I = 12/12.0 = 1.0 A
Self-mark with the mark scheme
Tick each mark point that your answer includes.
Self-mark the loop equation, combined drop and current.
Common mistake
Using only one resistor’s drop gives 3.0 A or 1.5 A and loses the second mark.
Check your understanding 2 Show the loop balance
For the 12 V, 4.0 Ω, 8.0 Ω series circuit, write the loop equation, find both resistor drops and verify that they add to the source rise.
Show answer
Check for 12-4.0I-8.0I = 0, I = 1.0 A, drops 4.0 V and 8.0 V and a total of 12 V. The checked current alone is only one part of this full method.
Guided practice 1
Single loop with internal resistance
Problem
A cell has e.m.f. ε = 1.5 V and internal resistance r = 0.50 Ω, connected to R = 2.0 Ω. Find the current.
Try this before viewing the solution
Show the steps Optional working
Hints
Hint 1
write every loop change
Traversing with the current gives + ε-Ir-IR = 0.
Show solution step by step
Account for both resistances
Method
ε-Ir-IR = 0.Reason
The source supplies energy per unit charge and both resistances dissipate it.
Working
Answer
ε = I(R + r)Calculate
Method
I = 0.60 A.Reason
The total series resistance is 2.50 Ω.Working
Answer
I = 1.5/(2.0 + 0.5) = 0.60 A
Quick check
The external drop is 1.20 V and the internal drop is 0.30 V; together they equal the 1.50 V e.m.f.
Try it yourself 1
Two opposing sources in series
Problem
Two sources of 9.0 V and 3.0 V oppose each other in a single loop with total resistance R = 6.0 Ω. Find the current magnitude and identify which source sets its direction.
Try this before viewing the solution
Show the steps Optional working
Hints
Hint 1
combine signed source changes
Choose the 9.0 V source direction as positive: ε net = 9.0-3.0.
Show solution step by step
Find the signed net e.m.f.
Method
ε net = 6.0 V in the direction of the 9.0 V source.
Reason
The two sources oppose, so their e.m.f.s subtract.
Working
Answer
9.0-3.0 = 6.0 VCalculate current magnitude
Method
I = 1.0 A.Reason
The net source rise is balanced by the drop across 6.0 Ω.
Working
Answer
I = 6.0/6.0 = 1.0 A
State the direction
Method
Current flows in the direction driven by the 9.0 V source.
Reason
The larger source determines the sign of the net e.m.f.
Working
Answer
ε net > 0 in the 9.0 V reference direction.
Quick check
Adding the opposing e.m.f.s would incorrectly treat the sources as aiding one another.
Circuits with more than one loop
- Label a current in each branch, with an assumed direction.
- Write the junction equations from Kirchhoff’s first law.
- Write a loop equation for each independent loop. Start with the smallest loops, and go round each in the direction of your assumed currents where possible.
- Solve the simultaneous equations. A negative current flows opposite to your arrow.
- Check by substituting the currents into a loop you did not use.
Worked example 1
Two sources feeding a shared resistor
Problem
Use Kirchhoff’s laws to find the branch currents in the circuit shown.
Assume I₁ and I₂ flow towards the junction and combine as I₃. Given R₁ = 10 Ω, R₂ = 20 Ω, R₃ = 40 Ω, and sources of 10 V and 20 V.
Show full solution
Write the junction relation
Method
I₁ + I₂ = I₃.Reason
The two assumed incoming branch currents combine in the shared resistor.
Working
Answer
I₃ = I₁ + I₂Write one equation for each independent loop
Method
Use each resistor’s own current arrow to decide its signed term.
Reason
Each closed path returns to the same potential.Working
Answer
10 = 10I₁ + 40I₃, 20 = 20I₂ + 40I₃Eliminate the shared current
Method
Substitute I₃ = I₁ + I₂ into both loop equations.
Reason
This leaves two simultaneous equations in I₁ and I₂.
Working
Answer
1 = 5I₁ + 4I₂, 1 = 2I₁ + 3I₂Solve the branch currents
Method
I₂ = 0.429 A and I₁ = -0.143 A.
Reason
Solving the two independent equations determines both signed currents.
Working
Answer
I₂ = 3/7 A, I₁ = -1/7 A
Recover and interpret the shared current
Method
I₃ = 0.286 A, while I₁ actually flows opposite to its assumed arrow.
Reason
I₃ = I₁ + I₂ and a negative signed current reverses the chosen reference direction.
Working
Answer
I₃ = -1/7 + 3/7 = 2/7 = 0.286 A
Quick check
Substituting the signed currents back into both loop equations reproduces the two source voltages.
Check your understanding 3 One source becomes stronger
In the two-source shared-resistor diagram above, replace only the 10 V source by a 30 V source with the same polarity. Keep R₁ = 10 Ω, R₂ = 20 Ω, R₃ = 40 Ω and the 20 V source. Use the same current arrows. Calculate all three signed currents; then write and solve the junction and two loop equations on paper.
Show the steps Optional working
Show answer
I₃ = I₁ + I₂, 30 = 10I₁ + 40I₃, 20 = 20I₂ + 40I₃. Substitution gives 3 = 5I₁ + 4I₂ and 1 = 2I₁ + 3I₂. Solving gives I₁ = 5/7 = 0.714 A, I₂ = -1/7 = -0.143 A and I₃ = 4/7 = 0.571 A. The stronger first source now drives current back through the second branch.
Check your understanding 4 Compare the complete network solution
Keep your full simultaneous-equation working on paper. Explain how you chose each resistor current in the loop equations, interpret the negative branch current and verify both source balances.
Show answer
Check that the shared resistor uses I₃, not I₁ or I₂ alone. Your equations should be I₃ = I₁ + I₂, 30 = 10I₁ + 40I₃ and 20 = 20I₂ + 40I₃. With the calculated signed currents, the first right-hand side is 30 V and the second is 20 V. I₂ flows opposite its assumed arrow. Reversing a loop traversal changes every term; it does not change the physical currents.
Mind stretcher 1 When changing magnetic flux mattersExtension
The loop equations in this lesson contain only source e.m.f.s and resistive drops. What else must you include if the magnetic flux through the loop is changing?
Show answer
A changing magnetic flux through the loop induces an e.m.f. (Faraday’s law), so that induced e.m.f. must appear in the loop equation. Leaving it out makes the model incomplete; energy is still conserved.
Common mistakes
- Changing the sign convention partway round a loop.
- Writing + IR for a resistor crossed in the direction of the current.
- Forgetting the internal resistance of a source when it is given.
- Adding the e.m.f.s of sources that oppose each other.
- Writing more loop equations than there are independent loops.
- Forgetting that a negative current flows opposite to the assumed direction.
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