Calculate the current in each branch of the (2025)

Calculate the current in each branch of the following circuit using mesh analysis?

C -- 20Ω -- D|30Ω 40ΩB -- H -- E|60Ω 50Ω 10ΩA -- G -- F100V 20V 50V

Ideas for Solving the Problem

1. Mesh Analysis: Mesh analysis (also known as loop analysis) is a method used to determine the currents flowing in a planar circuit (a circuit that can be drawn on a flat surface without any wires crossing). It involves assigning mesh currents to each independent loop in the circuit and then applying Kirchhoff's Voltage Law (KVL) to each loop.
2. Kirchhoff's Voltage Law (KVL): KVL states that the sum of the voltage drops around any closed loop in a circuit must equal zero.
3. Ohm's Law: Ohm's Law states that the voltage drop across a resistor is equal to the product of the current flowing through it and its resistance (V = IR).
4. Supermesh: When a current source is present between two meshes, we can form a supermesh by excluding the current source and any elements connected in series with it.

Calculation Steps

Step 1: Assign Mesh Currents

Assign mesh currents I1, I2, and I3 to the three independent loops in the circuit. Assume the currents flow clockwise.

Step 2: Apply KVL to Mesh 1 (ABCHA)

-60I1 - 30(I1 - I2) = -100
Simplifying:
90I1 - 30I2 = 100 (Equation 1)

Step 3: Apply KVL to Mesh 2 (BCDEH)

-30*(I2 - I1) - 20I2 - 40(I2 - I3) = 0
Simplifying:
30I1 + 90I2 - 40*I3 = 0 (Equation 2)

Step 4: Apply KVL to Mesh 3 (DEFGF)

-40*(I3 - I2) - 10I3 - 50I3 = -50 - 20
Simplifying:
-40I2 + 100I3 = 70 (Equation 3)

Step 5: Solve the System of Equations

We now have a system of three equations with three unknowns:

1. 90I1 - 30I2 = 100
2. 30I1 - 90I2 + 40*I3 = 0
3. -40I2 + 100I3 = 70

Solving this system of equations (using substitution, elimination, or a matrix solver), we get:
I1 ≈ 1.48 A
I2 ≈ 1.33 A
I3 ≈ 1.26 A

Step 6: Calculate Branch Currents

Now, we can calculate the current in each branch:

• Current in AB: I_AB = I1 ≈ 1.48 A
• Current in BC: I_BC = I1 - I2 ≈ 1.48 - 1.33 = 0.15 A
• Current in CD: I_CD = I2 ≈ 1.33 A
• Current in DE: I_DE = I2 - I3 ≈ 1.33 - 1.26 = 0.07 A
• Current in EF: I_EF = I3 ≈ 1.26 A
• Current in AG: I_AG = I1 ≈ 1.48 A
• Current in GF: I_GF = I3 ≈ 1.26 A
• Current in HG: I_HG = I2 - I3 ≈ 1.33 - 1.26 = 0.07 A

Final Answer

The branch currents are:

• I_AB ≈ 1.48 A
• I_BC ≈ 0.15 A
• I_CD ≈ 1.33 A
• I_DE ≈ 0.07 A
• I_EF ≈ 1.26 A
• I_AG ≈ 1.48 A
• I_GF ≈ 1.26 A
• I_HG ≈ 0.07 A

Highlights

• Mesh analysis is a powerful tool for solving complex circuits.
• Carefully define the mesh currents and apply KVL correctly.
• Pay attention to the direction of current flow when calculating branch currents.
• Double-check your calculations to avoid errors.
• Solving the system of equations can be done using various methods, including calculators or software.

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