Which point on Figure E5-1 best represents the impedance of a series circuit consisting of a 400-ohm resistor and a 38-picofarad capacitor at 14 MHz?

Q: Which point on Figure E5-1 best represents the impedance of a series circuit consisting of a 400-ohm resistor and a 38-picofarad capacitor at 14 MHz?

The correct answer is B: Point 4. The correct answer is B: Point 4. Which point on Figure E5-1 best represents the impedance of a series circuit consisting of a 400-ohm resistor and a 38-picofarad capacitor at 14 MHz is Point 4. At 14 MHz, XC = 1/(2πfC) ≈ 300Ω. Impedance ≈ 400 - j300, which corresponds to Point 4. For amateur radio operators, this is important for circuit analysis. Understanding this helps when reading impedance charts.

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Deep Dive: E5C10

The correct answer is B: Point 4. Which point on Figure E5-1 best represents the impedance of a series circuit consisting of a 400-ohm resistor and a 38-picofarad capacitor at 14 MHz is Point 4. At 14 MHz, XC = 1/(2πfC) ≈ 300Ω. Impedance ≈ 400 - j300, which corresponds to Point 4. For amateur radio operators, this is important for circuit analysis. Understanding this helps when reading impedance charts.

Why Other Answers Are Wrong

Option A (Point 2): Incorrect. Point 2 doesn't represent this impedance - Point 4 does. Point 2 is wrong. Option C (Point 5): Incorrect. Point 5 doesn't represent this impedance - Point 4 does. Point 5 is wrong. Option D (Point 6): Incorrect. Point 6 doesn't represent this impedance - Point 4 does. Point 6 is wrong.

Exam Tip

400Ω resistor + 38pF capacitor at 14 MHz = Point 4 on Figure E5-1. Think '4'00Ω + '3'8pF at '1'4 MHz = 'P'oint '4'. At 14 MHz, XC ≈ 300Ω, impedance ≈ 400 - j300, Point 4. Not Point 2, not Point 5, not Point 6 - just Point 4.

Memory Aid

400Ω resistor + 38pF capacitor at 14 MHz = Point 4 on Figure E5-1. Think '4'00Ω + '3'8pF = 'P'oint '4'. At 14 MHz, XC ≈ 300Ω, impedance ≈ 400 - j300, Point 4. Important for circuit analysis.

Real-World Example

A series circuit with 400-ohm resistor and 38-picofarad capacitor at 14 MHz: At 14 MHz, XC = 1/(2π×14×10^6×38×10^-12) ≈ 300Ω. Impedance ≈ 400 - j300 ohms. On Figure E5-1, this corresponds to Point 4. This is the point - Point 4.

Source & Coverage

Question Pool: 2024-2028 Question Pool

Subelement: E5C

Reference: 2024-2028 Question Pool · E5 - Electrical Principles

Key Concepts

Figure E5-1 Series circuit 400-ohm resistor 38-picofarad capacitor 14 MHz Point 4

Official Pool Source

Question text and answer key are synchronized with the current NCVEC Extra Class pool and mapped to the E5C topic.