Question

chapter 3, problem 54 given the circuit fig. a, use mesh analysis to find the mesh currents. figure 3.99 for prob. 3.54

Explanation:

Step1: Define Mesh Currents and Resistances

Let the mesh currents be \( i_1 \), \( i_2 \), and \( i_3 \) (all in mA for simplicity, since resistors are in kΩ). The resistors are \( R = 1\,\text{kΩ} \).

Step2: Write Mesh Equations for \( i_1 \)

For mesh 1: The voltage sources and resistors. The loop has \( 12\,\text{V} \) (positive to negative in the direction of \( i_1 \)), \( -10\,\text{V} \) (negative to positive against \( i_1 \)), and resistors: \( 1\,\text{kΩ} \) (shared with \( i_2 \), so \( i_1 - i_2 \) through it), and \( 1\,\text{kΩ} \) (only \( i_1 \)).
Equation: \( 1(i_1 - i_2) + 1i_1 + 12 - 10 = 0 \)
Simplify: \( 2i_1 - i_2 + 2 = 0 \) → \( 2i_1 - i_2 = -2 \) ...(1)

Step3: Write Mesh Equation for \( i_2 \)

For mesh 2: Resistors \( 1\,\text{kΩ} \) (from \( i_1 \) to \( i_2 \)), \( 1\,\text{kΩ} \) (shared with \( i_3 \), \( i_2 - i_3 \)), and \( 1\,\text{kΩ} \) (only \( i_2 \)), plus \( 10\,\text{V} \) (positive to negative in \( i_2 \) direction).
Equation: \( 1(i_2 - i_1) + 1(i_2 - i_3) + 1i_2 + 10 = 0 \)
Simplify: \( -i_1 + 3i_2 - i_3 + 10 = 0 \) → \( -i_1 + 3i_2 - i_3 = -10 \) ...(2)

Step4: Write Mesh Equation for \( i_3 \)

For mesh 3: Resistors \( 1\,\text{kΩ} \) (from \( i_2 \) to \( i_3 \)), \( 1\,\text{kΩ} \) (only \( i_3 \)), and \( -12\,\text{V} \) (negative to positive against \( i_3 \)).
Equation: \( 1(i_3 - i_2) + 1i_3 - 12 = 0 \)
Simplify: \( -i_2 + 2i_3 - 12 = 0 \) → \( -i_2 + 2i_3 = 12 \) ...(3)

Step5: Solve the System of Equations

From equation (1): \( i_2 = 2i_1 + 2 \)
Substitute \( i_2 \) into equation (2):
\( -i_1 + 3(2i_1 + 2) - i_3 = -10 \)
\( -i_1 + 6i_1 + 6 - i_3 = -10 \)
\( 5i_1 - i_3 = -16 \) ...(2a)

Substitute \( i_2 = 2i_1 + 2 \) into equation (3):
\( -(2i_1 + 2) + 2i_3 = 12 \)
\( -2i_1 - 2 + 2i_3 = 12 \)
\( -2i_1 + 2i_3 = 14 \) → \( -i_1 + i_3 = 7 \) ...(3a)

Now solve (2a) and (3a):
From (3a): \( i_3 = i_1 + 7 \)
Substitute into (2a):
\( 5i_1 - (i_1 + 7) = -16 \)
\( 4i_1 - 7 = -16 \)
\( 4i_1 = -9 \) → \( i_1 = -\frac{9}{4} = -2.25\,\text{mA} \)

Then \( i_3 = -2.25 + 7 = 4.75\,\text{mA} \)
And \( i_2 = 2(-2.25) + 2 = -4.5 + 2 = -2.5\,\text{mA} \)

Answer:

The mesh currents are \( i_1 = -2.25\,\text{mA} \), \( i_2 = -2.5\,\text{mA} \), \( i_3 = 4.75\,\text{mA} \) (or in A: \( i_1 = -2.25 \times 10^{-3}\,\text{A} \), \( i_2 = -2.5 \times 10^{-3}\,\text{A} \), \( i_3 = 4.75 \times 10^{-3}\,\text{A} \))