In a series RL circuit where R = 100 Ω and XL = 50 Ω. What is the total impedance of this circuit?
A) 111.8 $\Omega \angle$
26.6°
B) 111.8 $\Omega \angle$
-26.6°
C) 111.8 $\Omega \angle$
-63.4°
D) 111.8 $\Omega \angle$
63.4°

Question

In a series RL circuit where R = 100 Ω and XL = 50 Ω. What is the total impedance of this circuit?
A) 111.8 $\Omega \angle$
 26.6°
B) 111.8 $\Omega \angle$
 -26.6°
C) 111.8 $\Omega \angle$
 -63.4°
D) 111.8 $\Omega \angle$
 63.4°

Quizplus · Accepted Answer

The total impedance in a series RL circuit is calculated using Z = √(R² + XL²) and the phase angle θ = arctan(XL/R). Here, Z = √(100² + 50²) = 111.8 Ω and θ = arctan(50/100) = 26.6°.

In a Series RL Circuit Where R = 100 Ω Ω∠\Omega \angleΩ∠

In a Series RL Circuit Where R = 100 Ω $\Omega \angle$