1 Answer
Sure! Let's go through this **carefully and clearly**.
---
## What is Impedance?
**Impedance** (symbol: \( Z \)) is like resistance but for **alternating current (AC)** circuits. It includes both:
- **Resistance** (\( R \)): how much the component resists the current, and
- **Reactance** (\( X \)): how much the component opposes the change in current due to capacitors and inductors.
Impedance is **measured in ohms (Ω)**, just like resistance, but it is a **complex quantity**, meaning it has both a **real part** (resistance) and an **imaginary part** (reactance).
Mathematically:
\[
Z = R + jX
\]
where:
- \( R \) = resistance (in ohms)
- \( X \) = reactance (in ohms)
- \( j \) = imaginary unit ( \( j^2 = -1 \) ; engineers use \( j \) instead of \( i \)).
---
## Step-by-Step: How to Calculate Impedance
### 1. Identify the Components
First, check what components are in the circuit:
- **Resistor** — has only resistance (\( R \)).
- **Inductor** — has inductive reactance (\( X_L \)).
- **Capacitor** — has capacitive reactance (\( X_C \)).
Each component affects impedance differently.
---
### 2. Find the Reactance
Reactance is frequency-dependent. You must know the **AC frequency** (\( f \)).
- **Inductive Reactance** (\( X_L \)):
\[
X_L = 2\pi f L
\]
where:
- \( f \) = frequency in hertz (Hz)
- \( L \) = inductance in henrys (H)
- **Capacitive Reactance** (\( X_C \)):
\[
X_C = \frac{1}{2\pi f C}
\]
where:
- \( C \) = capacitance in farads (F)
> So, inductors "resist" high frequencies more, while capacitors "resist" low frequencies more.
---
### 3. Write Each Impedance
Each component's impedance:
- Resistor: \( Z_R = R \)
- Inductor: \( Z_L = jX_L \)
- Capacitor: \( Z_C = -jX_C \)
Notice inductors have **positive** imaginary impedance, capacitors have **negative** imaginary impedance.
---
### 4. Combine Impedances (Series or Parallel)
**Series Connection** (components one after another):
\[
Z_{\text{total}} = Z_1 + Z_2 + Z_3 + \cdots
\]
**Parallel Connection** (components connected side-by-side):
\[
\frac{1}{Z_{\text{total}}} = \frac{1}{Z_1} + \frac{1}{Z_2} + \frac{1}{Z_3} + \cdots
\]
When combining, remember to handle **complex numbers** properly!
---
### 5. Express Impedance
After combining, you often express the total impedance either:
- In **rectangular form**: \( Z = R + jX \), or
- In **polar form**:
\[
Z = |Z| \angle \theta
\]
where:
- \( |Z| = \sqrt{R^2 + X^2} \) is the magnitude
- \( \theta = \tan^{-1}\left(\frac{X}{R}\right) \) is the phase angle (in degrees or radians)
---
## Example Problem
**Given:**
- A resistor \( R = 10 \Omega \)
- An inductor \( L = 0.1 \, H \)
- A capacitor \( C = 100 \, \mu F = 100 \times 10^{-6} \, F \)
- AC frequency \( f = 50 \, Hz \)
**Find the total impedance if all are connected in series.**
---
**Step 1:** Find Reactances
- Inductive reactance:
\[
X_L = 2\pi f L = 2\pi \times 50 \times 0.1 = 31.42\, \Omega
\]
- Capacitive reactance:
\[
X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 100 \times 10^{-6}} = 31.83\, \Omega
\]
---
**Step 2:** Write Impedances
- Resistor: \( Z_R = 10 \, \Omega \)
- Inductor: \( Z_L = j31.42 \, \Omega \)
- Capacitor: \( Z_C = -j31.83 \, \Omega \)
---
**Step 3:** Add Them Up (Series)
\[
Z_{\text{total}} = Z_R + Z_L + Z_C
= 10 + j31.42 - j31.83
= 10 - j0.41 \, \Omega
\]
---
**Step 4:** Find Magnitude and Phase
- Magnitude:
\[
|Z| = \sqrt{10^2 + (-0.41)^2} \approx 10.008\, \Omega
\]
- Phase:
\[
\theta = \tan^{-1}\left(\frac{-0.41}{10}\right) \approx -2.35^\circ
\]
**Thus**, the total impedance is:
- In rectangular form: \( 10 - j0.41\, \Omega \)
- In polar form: \( 10.008 \angle -2.35^\circ \, \Omega \)
---
## Summary
✅ To calculate impedance:
1. Find resistance, inductive reactance, and capacitive reactance.
2. Assign proper impedance to each component.
3. Combine using series or parallel rules.
4. Express in rectangular or polar form as needed.
---
Would you also like me to show how this works for a purely **parallel** circuit too? (It looks a little different but is also very useful to know!) ✨
---
## What is Impedance?
**Impedance** (symbol: \( Z \)) is like resistance but for **alternating current (AC)** circuits. It includes both:
- **Resistance** (\( R \)): how much the component resists the current, and
- **Reactance** (\( X \)): how much the component opposes the change in current due to capacitors and inductors.
Impedance is **measured in ohms (Ω)**, just like resistance, but it is a **complex quantity**, meaning it has both a **real part** (resistance) and an **imaginary part** (reactance).
Mathematically:
\[
Z = R + jX
\]
where:
- \( R \) = resistance (in ohms)
- \( X \) = reactance (in ohms)
- \( j \) = imaginary unit ( \( j^2 = -1 \) ; engineers use \( j \) instead of \( i \)).
---
## Step-by-Step: How to Calculate Impedance
### 1. Identify the Components
First, check what components are in the circuit:
- **Resistor** — has only resistance (\( R \)).
- **Inductor** — has inductive reactance (\( X_L \)).
- **Capacitor** — has capacitive reactance (\( X_C \)).
Each component affects impedance differently.
---
### 2. Find the Reactance
Reactance is frequency-dependent. You must know the **AC frequency** (\( f \)).
- **Inductive Reactance** (\( X_L \)):
\[
X_L = 2\pi f L
\]
where:
- \( f \) = frequency in hertz (Hz)
- \( L \) = inductance in henrys (H)
- **Capacitive Reactance** (\( X_C \)):
\[
X_C = \frac{1}{2\pi f C}
\]
where:
- \( C \) = capacitance in farads (F)
> So, inductors "resist" high frequencies more, while capacitors "resist" low frequencies more.
---
### 3. Write Each Impedance
Each component's impedance:
- Resistor: \( Z_R = R \)
- Inductor: \( Z_L = jX_L \)
- Capacitor: \( Z_C = -jX_C \)
Notice inductors have **positive** imaginary impedance, capacitors have **negative** imaginary impedance.
---
### 4. Combine Impedances (Series or Parallel)
**Series Connection** (components one after another):
\[
Z_{\text{total}} = Z_1 + Z_2 + Z_3 + \cdots
\]
**Parallel Connection** (components connected side-by-side):
\[
\frac{1}{Z_{\text{total}}} = \frac{1}{Z_1} + \frac{1}{Z_2} + \frac{1}{Z_3} + \cdots
\]
When combining, remember to handle **complex numbers** properly!
---
### 5. Express Impedance
After combining, you often express the total impedance either:
- In **rectangular form**: \( Z = R + jX \), or
- In **polar form**:
\[
Z = |Z| \angle \theta
\]
where:
- \( |Z| = \sqrt{R^2 + X^2} \) is the magnitude
- \( \theta = \tan^{-1}\left(\frac{X}{R}\right) \) is the phase angle (in degrees or radians)
---
## Example Problem
**Given:**
- A resistor \( R = 10 \Omega \)
- An inductor \( L = 0.1 \, H \)
- A capacitor \( C = 100 \, \mu F = 100 \times 10^{-6} \, F \)
- AC frequency \( f = 50 \, Hz \)
**Find the total impedance if all are connected in series.**
---
**Step 1:** Find Reactances
- Inductive reactance:
\[
X_L = 2\pi f L = 2\pi \times 50 \times 0.1 = 31.42\, \Omega
\]
- Capacitive reactance:
\[
X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 100 \times 10^{-6}} = 31.83\, \Omega
\]
---
**Step 2:** Write Impedances
- Resistor: \( Z_R = 10 \, \Omega \)
- Inductor: \( Z_L = j31.42 \, \Omega \)
- Capacitor: \( Z_C = -j31.83 \, \Omega \)
---
**Step 3:** Add Them Up (Series)
\[
Z_{\text{total}} = Z_R + Z_L + Z_C
= 10 + j31.42 - j31.83
= 10 - j0.41 \, \Omega
\]
---
**Step 4:** Find Magnitude and Phase
- Magnitude:
\[
|Z| = \sqrt{10^2 + (-0.41)^2} \approx 10.008\, \Omega
\]
- Phase:
\[
\theta = \tan^{-1}\left(\frac{-0.41}{10}\right) \approx -2.35^\circ
\]
**Thus**, the total impedance is:
- In rectangular form: \( 10 - j0.41\, \Omega \)
- In polar form: \( 10.008 \angle -2.35^\circ \, \Omega \)
---
## Summary
✅ To calculate impedance:
1. Find resistance, inductive reactance, and capacitive reactance.
2. Assign proper impedance to each component.
3. Combine using series or parallel rules.
4. Express in rectangular or polar form as needed.
---
Would you also like me to show how this works for a purely **parallel** circuit too? (It looks a little different but is also very useful to know!) ✨