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Series RL Circuits 

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. V_{R} is still in phase with I, and V_{L} is still 90° out of phase with I. However, this time V_{L} 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 V_{AC} = 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 for this example circuit are shown to the right. This time the composite phase angle is positive instead of negative, because V_{L} leads I_{L}. But to determine just what that phase angle is, we must start by determining X_{L} and then calculating the rest of the circuit parameters.
X_{L}  =  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  
V_{R}  =  I × R 
=  0.0119 × 560  
=  6.656 V  
V_{L}  =  I × X_{L} 
=  0.0119 × 628  
=  7.464 V  
θ  =  arctan(X_{L}/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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