Answer:
a. 7 Ω
b. 5 Ω
c. 4.472 Ω
Step-by-step explanation:
a. Since we have two resistors in with resistances R = 4Ω and R' = 3 Ω in series, the total impedance z of the circuit is the sum of the individual resistances.
So, z = R + R' = 4 Ω + 3Ω = 7 Ω
b. Since we have a resistor of resistance R = 4Ω in series with an inductor of reactance X = 3 Ω, the total impedance is z = √(R² + X²)
z = √((4 Ω)² + (3 Ω)²)
z = √(16 Ω² + 9 Ω²)
z = √(25 Ω²)
z = 5 Ω
c. Since we have a resistor of resistance R = 4Ω in series with an inductor of reactance X = 3 Ω, and a capacitor of reactance, X' = 1 Ω the total impedance is z = √[R² + (X - X')²]
z = √((4 Ω)² + (3 Ω - 1 Ω)²)
z = √(16 Ω² + (2 Ω)²)
z = √(16 Ω² + 4 Ω²)
z = √(20 Ω²)
z = √(4 × 5) Ω
z = √4 × √5 Ω
z = 2√5 Ω
z = 4.472 Ω
