What is the total impedance of a series circuit with a resistor of 4Ω and an ind

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Question: What is the total impedance of a series circuit with a resistor of 4Ω and an inductor with a reactance of 3Ω?

Options:

Correct Answer: 7Ω

Solution:

In a series circuit, total impedance Z = R + jX = 4Ω + j3Ω, which gives a magnitude of √(4² + 3²) = 5Ω.

What is the total impedance of a series circuit with a resistor of 4Ω and an inductor with a reactance of 3Ω?

What is the total impedance of a series circuit with a resistor of 4Ω and an inductor with a reactance of 3Ω?

Step 1: Identify the resistance (R) in the circuit. Here, R = 4Ω.
Step 2: Identify the reactance (X) of the inductor in the circuit. Here, X = 3Ω.
Step 3: Write the formula for total impedance (Z) in a series circuit: Z = R + jX.
Step 4: Substitute the values of R and X into the formula: Z = 4Ω + j3Ω.
Step 5: To find the magnitude of the total impedance, use the formula: |Z| = √(R² + X²).
Step 6: Calculate R²: 4² = 16.
Step 7: Calculate X²: 3² = 9.
Step 8: Add R² and X²: 16 + 9 = 25.
Step 9: Take the square root of the sum: √25 = 5Ω.
Step 10: The total impedance of the circuit is 5Ω.

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