Current through inductor, equivalent impedance

Reading the figure

The source current Is is not written on the scan, so the answer is given in terms of Is. Elements (ac, reactances at the source frequency):

• Z₄ = 1 Ω resistor across the source
• Z₃ = inductor with reactance 5 Ω, so Z₃ = j5 Ω
• Z₂ = 4 Ω resistor in series with a capacitor of reactance 2 Ω, so Z₂ = 4 − j2 Ω
• Z₁ = 2 Ω resistor

Z₃ connects the source node to the node where Z₂ and Z₁ (in parallel) join the return line.

 +-----+------[Z3 = j5]------+-------+
 |     |                     |       |
(Is)  [Z4=1]              [Z2=4-j2] [Z1=2]
 |     |                     |       |
 +-----+---------------------+-------+

Step 1: Z₂ in parallel with Z₁

Z₁₂ = Z₁Z₂/(Z₁ + Z₂) = 2(4 − j2)/(6 − j2) = 1.4 − j0.2 Ω

Step 2: Add the inductor

Z_b = Z₃ + Z₁₂ = j5 + 1.4 − j0.2 = 1.4 + j4.8 Ω

Step 3: Equivalent impedance seen by the source

Z_eq = Z₄ ∥ Z_b = (1)(1.4 + j4.8)/(2.4 + j4.8)

Z_eq = 0.917 + j0.167 Ω = 0.932 ∠10.3° Ω

Step 4: Current through the inductor (current divider)

I_L = Is × Z₄/(Z₄ + Z_b) = Is × 1/(2.4 + j4.8)

I_L = Is × (0.0833 − j0.1667) = 0.186 Is ∠−63.4°

Example: for Is = 10∠0° A, I_L = 1.86 A ∠−63.4°.

Answer: Z_eq = 0.932 ∠10.3° Ω; I_L = 0.186 × Is ∠−63.4° (Is is not given in the scan).