A multi-loop bridge circuit
Solve a bridge circuit using Kirchhoff’s laws to find branch currents, V_ab, and the equivalent resistance (A Level Physics).
On this page
This is an enrichment worked example using Kirchhoff’s laws (not explicitly listed in the 9478 syllabus), useful for analysing non-trivial d.c. circuits.
Start with the model
A. Bridge circuit
A bridge circuit is a network with a “bridge” branch between two intermediate nodes (here the 1 Ω between a and b).
B. Kirchhoff’s laws
- Junction rule: ∑ Iᵢₙ = ∑ Iₒᵤₜ
- Loop rule: ∑ Δ V = 0
Key relationships
- Start by defining a small number of currents (I₁, I₂, I₃).
- Use KCL at the junctions to express branch currents like (I₁-I₃) and (I₂ + I₃).
- Write only independent loop equations (usually two “outer” loops + one “top”/“bridge” loop).
- A negative current means the actual direction is opposite to the arrow you assumed.
- Label currents on the bridge. 2) Write 2 loop equations + 1 more independent equation. 3) Solve simultaneous equations. 4) Interpret signs. 5) Use V = IR to get V_ab and R_eq.
How the model works
A. Reading the circuit and current labels
Use the labels shown in the diagram:
- I₁ flows down the top-left 1 Ω
- I₂ flows down the top-right 1 Ω
- I₃ flows from a to b through the middle 1 Ω
Then the bottom branch currents are set by junction rule:
- through bottom-left 1 Ω: I₁-I₃
- through bottom-right 2 Ω: I₂ + I₃
B. Writing the equations (with the diagram’s currents)
Left outer loop (source → top-left 1Ω → bottom-left 1Ω → back to source):
13 = (1)I₁ + (1)(I₁-I₃)
Right outer loop (source → top-right 1Ω → bottom-right 2Ω → back to source):
13 = (1)I₂ + (2)(I₂ + I₃)
Top loop (c → a → b → c):
0 = -I₁ - I₃ + I₂
Common mistakes
- Forgetting to use KCL to express the bottom-branch currents correctly.
- Using too many loop equations that are not independent.
- Treating a negative current as “wrong” instead of “direction is opposite”.
Using the model in questions
- Once you have I₁ and I₂, the supply current is Iₜₒₜₐₗ = I₁ + I₂.
- Equivalent resistance seen by the source: R_eq = V/Iₜₒₜₐₗ
- For the bridge branch: V_ab = I₃(1 Ω) = I₃ V
Worked examples
Worked example 1
Write the independent equations
Problem
Using the current labels in the diagram, write three independent equations for I₁, I₂, and I₃.
Worked solution
Express dependent branch currents
Method
The lower-left current is I₁-I₃ and the lower-right current is I₂ + I₃.
Reason
These expressions follow from the junction rule at nodes a and b.
Working
I_(lower - left) = I₁-I₃; I_(lower - right) = I₂ + I₃
Write the left outer loop
Method
13 = 2I₁-I₃.Reason
The 13 V source balances drops across the top-left and lower-left 1 Ω resistors.
Working
13 = (1)I₁ + (1)(I₁-I₃) = 2I₁-I₃Write the right outer loop
Method
13 = 3I₂ + 2I₃.Reason
The right loop includes the top-right 1 Ω and lower-right 2 Ω resistors.
Working
13 = (1)I₂ + (2)(I₂ + I₃) = 3I₂ + 2I₃Write the independent top loop
Method
I₂ = I₁ + I₃.Reason
The potential changes across the two upper resistors and bridge branch sum to zero.
Working
0 = -I₁-I₃ + I₂
Guided practice 2
Solve for I₁, I₂, and I₃
Problem
Solve 13 = 2I₁-I₃, 13 = 3I₂ + 2I₃, and I₂ = I₁ + I₃ for the three signed currents.
Try this before viewing the solution
Hints
Hint 1: eliminate one current
Substitute I₂ = I₁ + I₃ into the right-loop equation to obtain 3I₁ + 5I₃ = 13.
Show solution step by step
Eliminate I₂
Method
The right loop becomes 3I₁ + 5I₃ = 13.Reason
Substituting the top-loop relation reduces the system to two unknowns.
Working
13 = 3(I₁ + I₃) + 2I₃ = 3I₁ + 5I₃Solve with the left loop
Method
I₁ = 6.0 A and I₃ = -1.0 A.
Reason
The equations 2I₁-I₃ = 13 and 3I₁ + 5I₃ = 13 are independent.
Working
I₁ = 6.0 A; I₃ = -1.0 A
Recover I₂
Method
I₂ = 5.0 A.Reason
Use the top-loop relationship after solving the other two currents.
Working
I₂ = I₁ + I₃ = 6.0-1.0 = 5.0 A
Spot the mistake 3
Find the current in each resistor
Learner claim
Given I₁ = 6.0 A, I₂ = 5.0 A and I₃ = -1.0 A for the assumed a-to-b bridge arrow, a learner replaces I₃ by + 1.0 A in every branch expression. Diagnose this and find the lower-left and lower-right currents.
Try this before viewing the solution
Show solution step by step
Interpret, but retain, the sign
Method
The bridge current has magnitude 1.0 A from b to a.
Reason
The negative result reverses the assumed a-to-b arrow, but its signed value remains I₃ = -1.0 A in the defined equations.
Working
I₃ = -1.0 A for the a-to-b reference
Calculate the lower-left branch
Method
The lower-left current is 7 A.Reason
The b-to-a bridge current joins I₁ at node a.Working
I₁-I₃ = 6-(-1) = 7 ACalculate the lower-right branch
Method
The lower-right current is 4 A.Reason
One ampere leaves the I₂ stream through the bridge toward node a.
Working
I₂ + I₃ = 5 + (-1) = 4 A
Exam-style question 4
Find the equivalent resistance
Examination question
The 13 V source supplies the bridge through upper-branch currents I₁ = 6.0 A and I₂ = 5.0 A. Find the total supply current and equivalent resistance. [3 marks]
Try this before viewing the solution
Show solution step by step
Use the source junction
1 markMethod
Iₜₒₜₐₗ = 11 A.
Reason
The source current splits into the two upper branches.
Working
Iₜₒₜₐₗ = I₁ + I₂ = 6.0 + 5.0 = 11 A
Define equivalent resistance
1 markMethod
R_eq = V/Iₜₒₜₐₗ.
Reason
The entire bridge can be represented by the resistance drawing the same source current at 13 V.
Working
R_eq = 13/11Calculate
1 markMethod
R_eq = 1.2 Ω to two significant figures.
Reason
The value characterises the complete network at its source terminals.
Working
R_eq = 13/11 = 1.1818… Ω ≈ 1.2 Ω
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark the supply-current sum, equivalent-resistance relation and result.
Try it yourself 5
Find V_ab
Independent transfer
Try this before viewing the solution
Hints
Hint 1: keep the reference directions paired
Show solution step by step
Relate signed current and p.d.
Method
V_ab = I₃R for the a-to-b reference.Reason
The passive sign convention pairs current direction with the voltage drop direction.Working
Vₐ-V_b = I₃(1 Ω)Calculate the signed p.d.
Method
V_ab = -1.0 V.Reason
The actual current flows from b to a, so potential falls from b to a.Working
V_ab = (-1.0)(1) = -1.0 VInterpret the sign
Method
Node b is 1.0 V above node a.Reason
Vₐ-V_b < 0 means V_b > Vₐ.Working
V_b-Vₐ = +1.0 V
Try a different situation
Mind stretcher 1: Node-potential check for V_abExtension
Use node potentials to check V_ab quickly after you know I₁ and I₂.
Show answer
Take the bottom node d as 0 V and the top node c as + 13 V.
Then:
Vₐ = 13 - I₁(1 Ω) = 13-6 = 7 V
V_b = 13 - I₂(1 Ω) = 13-5 = 8 V
So:
V_ab = Vₐ-V_b = 7-8 = -1 V
Mind stretcher 2: When would the bridge current be zero?Extension
What condition on the resistor values would make the bridge branch current I₃ equal to zero (a balanced bridge)? Explain briefly.
Show answer
A bridge is balanced when the two potential divider ratios match, so the potentials at nodes a and b are equal and no current flows through the bridge resistor.
In general: (R_(top-left))/(R_(bottom-left)) = (R_(top-right))/(R_(bottom-right)) (ratio condition).
Mind stretcher 3: Optional extensionExtension
A. When you can avoid Kirchhoff (bridge symmetry)
Some bridge circuits can be simplified without Kirchhoff’s laws using symmetry (e.g. balanced bridges). This example is unbalanced, so Kirchhoff’s laws are the most direct method.
Syllabus and review details
No official syllabus alignment is listed for this lesson.