In a series RLC circuit, R = 1.1 kΩ, XL = 1.6 kΩ, and XC = 2.9 kΩ. ZT = .
A) 4.6 k$\Omega \angle$
-50°
B) 4.6 k$\Omega \angle$
50°
C) 1.7 k$\Omega \angle$
50°
D) 1.7 k$\Omega \angle$
-50°

Question

In a series RLC circuit, R = 1.1 kΩ, XL = 1.6 kΩ, and XC = 2.9 kΩ. ZT = .
A) 4.6 k$\Omega \angle$
 -50°
B) 4.6 k$\Omega \angle$
 50°
C) 1.7 k$\Omega \angle$
 50°
D) 1.7 k$\Omega \angle$
 -50°

Quizplus · Accepted Answer

The impedance ZT of a series RLC circuit is calculated using ZT = √(R² + (XL - XC)²). Here, ZT = √(1.1² + (1.6 - 2.9)²) kΩ = √(1.1² + (-1.3)²) kΩ = √(1.21 + 1.69) kΩ = √2.9 kΩ ≈ 1.7 kΩ. The phase angle θ = arctan((XL - XC)/R) = arctan((-1.3)/1.1) ≈ -50°. Thus, ZT = 1.7 kΩ ∠ -50°.

In a Series RLC Circuit, R = 1 Ω∠\Omega \angleΩ∠ -50° B) 4

In a Series RLC Circuit, R = 1 $\Omega \angle$ -50°
B) 4