The given circuit is suitable for Mesh Current Analysis (also called Mesh Analysis or Loop Current Method).
Why it’s suitable:
- The circuit is a planar circuit (it can be drawn on a plane without crossing wires).
- It has closed loops, which is required for applying mesh analysis.
- There are multiple voltage sources and resistors, making it ideal for setting up mesh (loop) equations using Kirchhoff’s Voltage Law (KVL).
What to do:
- Assign mesh currents to each of the independent loops (clockwise or counterclockwise).
- Apply KVL to each loop: sum of voltage drops = sum of voltage sources.
- Solve the resulting system of equations to find the unknown mesh currents.
- Use the currents to find voltages or unknown resistance R if required.
1. Number of Mesh Loops
There are 4 mesh loops:
- Mesh 1 → Top-left
- Mesh 2 → Top-right
- Mesh 3 → Bottom-right
- Mesh 4 → Bottom-left
Mesh currents:
- I1 = mesh current in loop 1 (anticlockwise)
- I2 = mesh current in loop 2 (anticlockwise)
- I3 = mesh current in loop 3 (anticlockwise)
- I4 = mesh current in loop 4 (anticlockwise)
2. Resistor Polarity
Follow passive sign convention:
- Voltage drop across resistors is positive in the direction of mesh current.
- Shared resistors use current differences like (I1 - I2), etc.
3. KVL Equations
Mesh 1 (I1):
Loop contains: 20V source, 40Ω, 50Ω shared with I2
KVL:
-20 + 40I1 + 50(I1 - I2) = 0
Simplify:
90I1 - 50I2 = 20 → Equation (1)
Mesh 2 (I2):
Loop contains: 50Ω shared with I1, 30Ω shared with I3, 10V source
KVL:
50*(I2 - I1) + 30*(I2 - I3) + 10 = 0
Simplify:
-50I1 + 80I2 - 30*I3 = -10 → Equation (2)
Mesh 3 (I3):
Loop contains: 30Ω shared with I2, 20Ω, 30V source
KVL:
30*(I3 - I2) + 20I3 - 30 = 0
Simplify:
-30I2 + 50*I3 = 30 → Equation (3)
Mesh 4 (I4):
Loop contains: 50V source, 30Ω shared with I1, unknown resistor R
KVL:
-50 + 30*(I4 - I1) + RI4 = 0
Simplify:
-30I1 + (30 + R)*I4 = 50 → Equation (4)
Mesh Current Solution:
(1) 90I1 - 50I2 = 20
(2) -50I1 + 80I2 - 30I3 = -10
(3) -30I2 + 50I3 = 30
(4) -30I1 + (30 + R)I4 = 50 → Let R = 30, so it becomes:
-30I1 + 60*I4 = 50
Step 1: Solve Equation (3) for I3
From Equation (3):
-30I2 + 50I3 = 30
→ I3 = (30 + 30*I2) / 50
Step 2: Plug I3 into Equation (2)
-50I1 + 80I2 - 30*[(30 + 30I2)/50] = -10
→ -50I1 + 80I2 - (900 + 900I2)/50 = -10
→ -50I1 + 80I2 - 18 - 18I2 = -10
→ -50I1 + 62*I2 = 8 → Equation (2 simplified)
Step 3: Now solve Equations (1) and simplified (2)
From Equation (1):
90I1 - 50I2 = 20 → Multiply both sides by 5:
450I1 - 250I2 = 100
From simplified (2):
-50I1 + 62I2 = 8 → Multiply both sides by 9:
-450I1 + 558I2 = 72
Now add both equations:
(450I1 - 250I2) + (-450I1 + 558I2) = 100 + 72
→ 308*I2 = 172
→ I2 ≈ 0.558 A
Step 4: Plug I2 into Equation (1) to get I1
90I1 - 50I2 = 20
→ 90I1 - 50(0.558) = 20
→ 90I1 - 27.9 = 20
→ 90I1 = 47.9
→ I1 ≈ 0.532 A
Step 5: Plug I2 into Equation (3) to get I3
I3 = (30 + 30I2) / 50
→ I3 = (30 + 300.558) / 50
→ I3 = (30 + 16.74) / 50 = 46.74 / 50
→ I3 ≈ 0.935 A
Step 6: Plug I1 into Equation (4) to get I4
-30I1 + 60I4 = 50
→ -300.532 + 60I4 = 50
→ -15.96 + 60I4 = 50
→ 60I4 = 65.96
→ I4 ≈ 1.099 A
Final Results (Mesh Currents):
- I1 ≈ 0.532 A
- I2 ≈ 0.558 A
- I3 ≈ 0.935 A
- I4 ≈ 1.099 A