Understanding Phase Change: R, L & C

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Phase change is crucial in electrical systems because it highlights the differences in voltage and current across resistors (R), inductors (L), and capacitors (C). Complex impedances inherently cause a phase shift between the applied voltage and the resulting current. Specifically, in an inductor, the current lags behind the voltage, illustrating this phase relationship. Understanding these phase shifts is essential for accurately analyzing complex impedance problems. Thus, recognizing phase change is fundamental for effective circuit analysis and design.
hidemi
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Homework Statement
If the input to an RLC series circuit is V = Vm Cos ωt, then the current in the circuit is?
The answer is (D) as attached.
Relevant Equations
V = IZ
Z = [ R^2 + (XL - Xc)^2]^1/2
Why do we need to consider phase change?
Here are my thoughts: is it because voltages are different in phase for each of the three electrical accessories, R, L and C?
 

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Because complex impedances will, in general, cause a phase shift between the voltage applied and the current. It's what they do. Suppose Z is just an inductor, would the current be in phase with the voltage?
 
DaveE said:
Because complex impedances will, in general, cause a phase shift between the voltage applied and the current. It's what they do. Suppose Z is just an inductor, would the current be in phase with the voltage?
No, the current will be lagged after voltage.
 
hidemi said:
No, the current will be lagged after voltage.
Which will appear as a phase shift between the voltage and current. That is why phase matters for complex impedance problems.
 
DaveE said:
Which will appear as a phase shift between the voltage and current. That is why phase matters for complex impedance problems.
Thank you for your further explanation.
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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