---
title: "Current through inductor, equivalent impedance"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:current-through-inductor-equivalent-impedance"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/current-through-inductor-equivalent-impedance
site: BCT Engineering
---

# Current through inductor, equivalent impedance

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2072 Kartik, 6 marks
- Video: [How to find impedance and current in AC circuit ? | Impedance and Admittance | Electrical Engineer](https://www.youtube.com/watch?v=BSiz4Yb4Nzc) — Electrical and Electronics Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/current-through-inductor-equivalent-impedance
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

In the given circuit, find the current through the inductor, what is the equivalent impedance? [Figure: current source Is feeding a network with Z4 = 1 Ω shunt; Z3 = 5 Ω inductor on top; Z2 = 4 Ω in series with 2 Ω in the middle branch; Z1 = 2 Ω on the right.]

## Short answer

Z1 in parallel with Z2 yields 1.4-j0.2 Ω. Adding Z3=j5 in series gives Zb=1.4+j4.8 Ω. Equivalent impedance Z_eq = 1 Ω || Zb = 0.932∠10.3° Ω. Inductor current I_L = Is × 1/(Z4+Zb) = 0.186Is ∠-63.4° (Is not given).

## Answer

#### 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.

```text
 +-----+------[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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Z2 = 4 Ω in series with a capacitor of reactance 2 Ω is
   A. 4 + j2 Ω
   B. 6 Ω
   C. j8 Ω
   D. 4 - j2 Ω
   Answer: D. A capacitor has negative reactance, so Z2 = 4 - j2. (Key point: Z2 value)

2. Z1 (2 Ω) in parallel with Z2 gives
   A. 6 - j2 Ω
   B. 1.4 + j0.2 Ω
   C. 1.4 - j0.2 Ω
   D. 2 Ω
   Answer: C. Z1Z2/(Z1+Z2) = 2(4 - j2)/(6 - j2) = 1.4 - j0.2. (Key point: Parallel part)

3. The inductor branch (Z3 in series with the parallel pair) is
   A. j5 Ω
   B. 1.4 + j4.8 Ω
   C. 1.4 - j4.8 Ω
   D. 6.4 Ω
   Answer: B. j5 + 1.4 - j0.2 = 1.4 + j4.8. (Key point: Inductor branch)

4. The inductor current is found by
   A. I_L = Is x Z4/(Z4 + Zb), giving 0.186 Is at -63.4°
   B. I_L = Is x Zb/(Z4 + Zb)
   C. I_L = Is/Z3
   D. I_L = Is x Z4 x Zb
   Answer: A. Z4 is in parallel with the branch Zb, so current divides in proportion to the other branch's impedance. (Key point: Current divider)

## Related in Basic Electrical Engineering

- [Series-parallel circuit with I1, I2, I3 phasors](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-ac-v100-z1-z2-z3) — Basic Electrical Engineering, chapter 6, asked 2×
- [Current from source in ladder circuit](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asource-current-3v-2ohm-ladder) — Basic Electrical Engineering, chapter 2
- [Loop current method with 3 A source](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aloop-current-2-ohm-3a-9v-12v) — Basic Electrical Engineering, chapter 2
- [Source current from parallel branch currents](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aparallel-branch-currents-time-expression) — Basic Electrical Engineering, chapter 5
- [Nodal analysis, current through 15Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-current-15-ohm-8a-48v) — Basic Electrical Engineering, chapter 2
- [Nodal analysis current through 10Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-current-through-10-ohm-5a) — Basic Electrical Engineering, chapter 2, asked 2×

## Related in other subjects

- [Equivalent Z0, Z1, Z2 at bus 3](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Asequence-thevenin-z0-z1-z2-bus3) — Power System Analysis II, chapter 5
- [Laplace: current through 2 Ω, exponential source](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Alaplace-current-through-2ohm-exponential-source) — Electrical Circuit Theory, chapter 4
- [Classical: inductor current and capacitor voltage, 100 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-inductor-current-capacitor-voltage-100v) — Electrical Circuit Theory, chapter 3
- [State equations of current-source RLC circuit](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ass-current-source-1h-1f-2h) — Control Systems, chapter 8