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Series RL Circuits

The Circuit

Applying an ac voltage to an RL series circuit.

When we apply an ac voltage to a series RL circuit as shown to the right, the circuit behaves in some ways the same as the series RC circuit, and in some ways as a sort of mirror image. For example, current is still the same everywhere in this series circuit. VR is still in phase with I, and VL is still 90° out of phase with I. However, this time VL leads I — it is at +90° instead of -90°.

For this circuit, we will assign experimental values as follows: R = 560Ω, L = 100 mH, and VAC = 10 Vrms. We build the circuit, and measure 7.464 V across L, and 6.656 V across R. As we might have expected, this exceeds the source voltage by a substantial amount, and the phase shift is the reason for it.



The Vectors in an LC Series Circuit

The voltage vectors in a series RL circuit.

The vectors for this example circuit are shown to the right. This time the composite phase angle is positive instead of negative, because VL leads IL. But to determine just what that phase angle is, we must start by determining XL and then calculating the rest of the circuit parameters.

XL  =  2πfL
 
   =  6.28 × 1000 × 0.1
 
   =  628
 
 
Z  =  560 + j628 Ω
 
   =  (560² + 628²)½
 
   =  (313600 + 394384)½
 
   =  (707984)½
 
   =  841.4 Ω
 
 
I  =  E/Z
 
   =  10/841.4
 
   =  0.0119 A
 
   =  11.9 mA
 
 
VR  =  I × R
 
   =  0.0119 × 560
 
   =  6.656 V
 
 
VL  =  I × XL
 
   =  0.0119 × 628
 
   =  7.464 V
 
 
θ  =  arctan(XL/R)
 
   =  arctan(628/560)
 
   =  arctan(1.121)
 
   =  48.28°


This really completes the description of the series RL circuit with a fixed AC signal applied to it. Starting with the component values and the frequency of the applied AC voltage, we have described every aspect of this circuit's behavior at that frequency.


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