Mesh analysis: practice problems with step-by-step solutions
Also called: mesh current method · loop analysis · KVL
Free Mesh analysis practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: 2 meshes, 1 source, 2 meshes, 2 sources, 3 meshes.
Free practice problemEasy
Use mesh analysis to find the current IR1 through R1 in the direction of the arrow.
Mesh analysis is a standard method to find the currents in a circuit. It is the partner of nodal analysis: nodal analysis uses voltages as unknowns, mesh analysis uses currents.
Look at a flat circuit drawing. The empty “windows” enclosed by wires and parts are called meshes.
You imagine one current circling around each window. These are the mesh currents. Every real branch current is built from them.
Think of a row of gear wheels side by side. Each wheel turns on its own, and where two wheels touch, what you see is the difference between their motions.
Self-check A flat circuit drawing has three windows. How many mesh currents do you need?
1
2
3
6
Answer: 3. One mesh current per window, so three. You then write one equation per mesh, three equations in total.
2Kirchhoff’s voltage law
Walk once around any closed loop and come back to where you started. You end at the same voltage as you began.
So the voltage you gain in sources equals the voltage you lose in resistors. This is Kirchhoff’s voltage law (KVL).
Σ voltage rises = Σ voltage drops
A source is a rise when you pass through it from − to +. A resistor is a drop of I·R when you walk through it in the direction of its current.
In this loop the source gives 12 V. One resistor takes 8 V, so the other one must take the remaining 4 V.
Self-check A 10 V source drives one loop with two resistors. The first has 7 V across it. What is the voltage across the second?
10 V
17 V
3 V
7 V
Answer: 3 V. The drops must add up to the 10 V rise: 10 − 7 = 3 V. Adding the voltages (17 V) breaks KVL.
3Mesh currents in a shared resistor
Draw every mesh current clockwise, with the same name pattern: I1, I2, …
A resistor on the outside of the circuit belongs to one mesh only. It carries that mesh current.
A resistor between two windows is shared. Both mesh currents pass through it, in opposite directions. Its real current is the difference.
IR3 (down) = I1 − I2
In mesh 1, I1 flows down through R3, so you write I1 − I2. In mesh 2, I2 flows up through R3, so you write I2 − I1.
Self-check In the circuit above, I1 = 3 A and I2 = 1 A. What current flows down through R3?
1 A
2 A
3 A
4 A
Answer: 2 A. The shared resistor carries the difference: 3 − 1 = 2 A downwards. Adding the mesh currents (4 A) is the most common slip.
4The method step by step
Count the windows. Draw one clockwise mesh current in each: I1, I2, …
Go around mesh 1 clockwise. Add up the drops: for each resistor, R times the current through it (its own mesh current, or the difference in a shared resistor).
Set the drops equal to the source rises in that mesh. A source counts as + when you pass it from − to +, and as − the other way.
Repeat for every mesh. You get one equation per mesh.
Solve the equations for the mesh currents.
Find the branch current or voltage you need. A negative mesh current flows anticlockwise.
Always go around in the same direction as the mesh current. Then every resistor term starts with a plus sign.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
Find the mesh currents and the current IR3 through R3, in the direction of the arrow.
1
Assign mesh currents
There are two windows, so there are two clockwise mesh currents, I1 and I2. R3 is shared between them.
2
KVL around mesh 1
Going clockwise, you pass V1 from − to +: a rise of 10 V. R1 carries I1, and R3 carries I1 − I2.
2·I1 + 6·(I1 − I2) = 10
8·I1 − 6·I2 = 10
3
KVL around mesh 2
Mesh 2 has no source, so its drops add up to 0. R2 and R4 carry I2, and R3 carries I2 − I1.
4·I2 + 2·I2 + 6·(I2 − I1) = 0
−6·I1 + 12·I2 = 0 ⇒ I1 = 2·I2
4
Solve the equations
Put I1 = 2·I2 into the first equation.
8·2·I2 − 6·I2 = 10 ⇒ 10·I2 = 10
I2 = 1 A, I1 = 2 A
5
Current in R₃
The arrow points down, the same way as I1 in R3.
IR3 = I1 − I2 = 2 − 1 = 1 A
6
Check the answer
Mesh 1: R1 takes 2·2 = 4 V and R3 takes 6·1 = 6 V. Together 10 V, equal to V1.
Mesh 2: R2 and R4 carry 1 A and take 4 + 2 = 6 V. That equals the 6 V across R3, as it must, because they sit across the same two nodes.
Answer
I1 = 2 A, I2 = 1 A and IR3 = 1 A downwards.
6Common mistakes
Using the full mesh current in a shared resistor. A shared resistor carries the difference of two mesh currents.
Getting the sign of a source wrong. Passing a source from + to − in the direction of travel counts as a negative rise.
Going around some meshes anticlockwise. Keep every mesh current and every walk clockwise.
Forgetting that a negative result is allowed. It only means the current flows anticlockwise.
Self-check Going clockwise around mesh 2, you pass a 6 V source from its + side to its − side. Mesh 2 contains R2 = 2 Ω and the shared R3 = 4 Ω. Which equation is right?
2·I2 + 4·(I1 − I2) = −6
2·I2 + 4·(I2 − I1) = 6
2·I2 + 4·(I2 − I1) = −6
2·I2 + 4·I2 = −6
Answer: 2·I2 + 4·(I2 − I1) = −6. Passing a source from + to − is a drop, so it counts as −6 V on the rise side. In mesh 2 the shared resistor carries I2 − I1, not I2 alone and not I1 − I2.
7How exam questions look
Easy: two meshes and one voltage source, as in the worked example. Mesh 2 has no source, so its equation equals 0.
Medium: two meshes and two voltage sources, one on each side. Watch the sign of the right-hand source in mesh 2.
Hard: three meshes, with two shared resistors. The middle mesh shares a resistor with each neighbour.
The question asks for a mesh current, the current through a resistor in the direction of the arrow, or the voltage across a resistor with the polarity shown.
For a voltage, find the current through the resistor first and then use V = I·R.
Use mesh analysis to find the current IR2 through R2 in the direction of the arrow.
V1 = 10 VR1 = 8 ΩR2 = 8 ΩR3 = 3 ΩR4 = 8 Ω
Find IR2(A)
Show hints
There are 2 windows, so define 2 clockwise mesh currents (I1, I2) and write one KVL equation per mesh.
In mesh 1, every resistor carries I1, except a shared resistor, which carries I1 − I2.
Solving gives I1 = 0.95 A.
Show the answer and the step-by-step solution
Mesh analysis: give each window of a planar circuit its own loop current, write Kirchhoff’s voltage law around each loop (sum of drops = sum of rises), and solve. Branch currents are the mesh current, or the difference of two mesh currents on a shared branch.
Assign mesh currents
Draw a clockwise current in each window: I1, I2. A resistor shared by two meshes carries the difference of their currents.
KVL around mesh 1
Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).
R1·I1 + R3(I1 − I2) = V1
8·I1 + 3(I1 − I2) = 10
11·I1 − 3·I2 = 10
KVL around mesh 2
Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).
R2·I2 + R3(I2 − I1) + R4·I2 = 0
8·I2 + 3(I2 − I1) + 8·I2 = 0
−3·I1 + 19·I2 = 0
Solve
Δ = (11)(19) − (−3)(−3) = 200
I1 = Δ1/Δ = 190/200 = 0.95 A
I2 = Δ2/Δ = 30/200 = 0.15 A
Answer
IR2 = I2 = 0.15 A
A negative mesh current flows anticlockwise.
Example 2Medium
Use mesh analysis to find the mesh current I1 (clockwise).
V1 = 15 VV2 = 24 VR1 = 10 ΩR2 = 3 ΩR3 = 12 Ω
Find I1(A)
Show hints
There are 2 windows, so define 2 clockwise mesh currents (I1, I2) and write one KVL equation per mesh.
In mesh 1, every resistor carries I1, except a shared resistor, which carries I1 − I2.
Solving gives I1 = −0.3387 A.
Show the answer and the step-by-step solution
Mesh analysis: give each window of a planar circuit its own loop current, write Kirchhoff’s voltage law around each loop (sum of drops = sum of rises), and solve. Branch currents are the mesh current, or the difference of two mesh currents on a shared branch.
Assign mesh currents
Draw a clockwise current in each window: I1, I2. A resistor shared by two meshes carries the difference of their currents.
KVL around mesh 1
Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).
R1·I1 + R3(I1 − I2) = V1
10·I1 + 12(I1 − I2) = 15
22·I1 − 12·I2 = 15
KVL around mesh 2
Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).
R2·I2 + R3(I2 − I1) = −V2
3·I2 + 12(I2 − I1) = (−24)
−12·I1 + 15·I2 = −24
Solve
Δ = (22)(15) − (−12)(−12) = 186
I1 = Δ1/Δ = −63/186 = −0.3387 A
I2 = Δ2/Δ = −348/186 = −1.871 A
Answer
I1 = −0.3387 A
A negative mesh current flows anticlockwise.
Example 3Hard
Use mesh analysis to find the current IR4 through R4 in the direction of the arrow.
There are 3 windows, so define 3 clockwise mesh currents (I1, I2, I3) and write one KVL equation per mesh.
In mesh 1, every resistor carries I1, except a shared resistor, which carries I1 − I2.
Solving gives I1 = 1.135 A.
Show the answer and the step-by-step solution
Mesh analysis: give each window of a planar circuit its own loop current, write Kirchhoff’s voltage law around each loop (sum of drops = sum of rises), and solve. Branch currents are the mesh current, or the difference of two mesh currents on a shared branch.
Assign mesh currents
Draw a clockwise current in each window: I1, I2, I3. A resistor shared by two meshes carries the difference of their currents.
KVL around mesh 1
Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).
R1·I1 + R4(I1 − I2) = V1
15·I1 + 30(I1 − I2) = 36
45·I1 − 30·I2 = 36
KVL around mesh 2
Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).
R2·I2 + R4(I2 − I1) + R5(I2 − I3) = 0
20·I2 + 30(I2 − I1) + 8(I2 − I3) = 0
−30·I1 + 58·I2 − 8·I3 = 0
KVL around mesh 3
Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).
R3·I3 + R5(I3 − I2) = −V2
10·I3 + 8(I3 − I2) = (−15)
−8·I2 + 18·I3 = −15
Solve
Δ = det of the coefficient matrix = 27900
I1 = Δ1/Δ = 31680/27900 = 1.135 A
I2 = Δ2/Δ = 14040/27900 = 0.5032 A
I3 = Δ3/Δ = −17010/27900 = −0.6097 A
Answer
IR4 = I1 − I2 = 0.6323 A
A negative mesh current flows anticlockwise.
Questions and answers
What is Mesh analysis?
Mesh analysis is a standard method to find the currents in a circuit. It is the partner of nodal analysis: nodal analysis uses voltages as unknowns, mesh analysis uses currents.
Look at a flat circuit drawing. The empty “windows” enclosed by wires and parts are called meshes.
How do I solve Mesh analysis problems step by step?
Draw a clockwise current in every mesh (window) of the circuit.
Apply Kirchhoff’s voltage law around each mesh: the sum of voltage drops is zero.
In a resistor shared by two meshes the current is the difference of the two mesh currents.
Solve the equations; a negative result means the current flows the other way.
What are common mistakes in Mesh analysis?
Using the full mesh current in a shared resistor.
Getting the sign of a source wrong when going around the loop.
Forgetting that a current source in an outer branch fixes that mesh current.
What do the Mesh analysis exercises cover?
Easy: 2 meshes, 1 source
Medium: 2 meshes, 2 sources
Hard: 3 meshes
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.