Question

chapter 3, problem x given the circuit below, use mesh analysis to find the mesh currents voltages v₁, v₂ and v₃.

Explanation:

Step1: Write mesh equations

For mesh 1 (current \(i_1\)): The resistors in mesh 1 are \(2\Omega\), \(4\Omega\), and the voltage source is \(10V\). The current through \(4\Omega\) is \(i_1 - i_2\) (since \(i_1\) and \(i_2\) are in opposite directions through \(4\Omega\)). So the KVL equation is:

$$10 = 2i_1 + 4(i_1 - i_2)+ 6(i_1 - i_2)$$

Simplify: \(10 = 2i_1 + 4i_1 - 4i_2 + 6i_1 - 6i_2\) → \(10 = 12i_1 - 10i_2\) (Equation 1)

For mesh 2 (current \(i_2\)): The resistors are \(4\Omega\), \(6\Omega\), \(8\Omega\). The current through \(4\Omega\) is \(i_2 - i_1\), through \(6\Omega\) is \(i_2 - i_1\). So KVL equation:

$$0 = 4(i_2 - i_1)+ 6(i_2 - i_1)+ 8i_2$$

Simplify: \(0 = 4i_2 - 4i_1 + 6i_2 - 6i_1 + 8i_2\) → \(0 = -10i_1 + 18i_2\) (Equation 2)

Step2: Solve the system of equations

From Equation 2: \(10i_1 = 18i_2\) → \(i_1 = \frac{18}{10}i_2 = 1.8i_2\)

Substitute \(i_1 = 1.8i_2\) into Equation 1:

$$10 = 12(1.8i_2)- 10i_2$$

$$10 = 21.6i_2 - 10i_2$$

$$10 = 11.6i_2$$

$$i_2 = \frac{10}{11.6} \approx 0.862A$$

Then \(i_1 = 1.8\times0.862 \approx 1.552A\)

Step3: Find voltages \(v_1\), \(v_2\), \(v_3\) (assuming \(v_1\) is across \(2\Omega\), \(v_2\) across \(4\Omega\), \(v_3\) across \(8\Omega\))

\(v_1 = 2i_1 = 2\times1.552 \approx 3.104V\)

\(v_2 = 4|i_1 - i_2| = 4\times(1.552 - 0.862)= 4\times0.69 \approx 2.76V\)

\(v_3 = 8i_2 = 8\times0.862 \approx 6.896V\)

(Note: If \(v_1\), \(v_2\), \(v_3\) are across other resistors, adjust accordingly. The problem's \(v_1\), \(v_2\), \(v_3\) labels are not clear, but assuming standard labels: \(v_1\) (2Ω), \(v_2\) (4Ω), \(v_3\) (8Ω))

Answer:

Mesh currents: \(i_1 \approx 1.55A\), \(i_2 \approx 0.86A\)
Voltages (example): \(v_1 \approx 3.10V\), \(v_2 \approx 2.76V\), \(v_3 \approx 6.90V\)

(If specific resistor for \(v_1\), \(v_2\), \(v_3\) is different, recalculate. The key is solving mesh currents first via KVL and then voltages via Ohm's law \(V = IR\))