Construct the network of Fig 141 and insert the measured val

Construct the network of Fig. 14.1 and insert the measured values of the resistors in the spaces provided. Using branch-current analysis, calculate the current through each branch of the network of Fig. 14.1 and insert in Table 14.1. Use the measured resistor values and assume the current directions shown in the figure. Show all your calculations in the space provided and be neat!

Solution

-1200 I₁ – 3300(I ₁ + I ₂ ) + 20 = 0 here I1+I2= I3

- 2200I₂ - 3300 (I₁ + I₂ )+ 10= 0

-4500 I₁ - 3300I₂ + 20 = 0

-3300 I₁   - 5500 I₂ +10 =0

I₁ = -5.55mA (current through 1.2kohms)

I₂   = 1.51mA (current through 2.2kohms)

I₃ = -4.04mA (current through 3.3kohms)

actual current directions are like I2 remains same where I1 and I3 reverses