Answer:
a.223.6Ω
b.0.134A
c. [tex]V_{R}=26.8v[/tex] , [text]V_{L}= 13.4V[/tex]
d.63.43, the voltage lead the current
e. see attachment
Explanation:
the diagram can be shown in the attached document.
We can write the voltage as a function of time
[tex]v(t)=Vsin(wt+\alpha )\\[/tex]
where V= amplitude of the voltage
w=angular frequency.
from the question we have V=30v and w=250rad/s
hence we can rewrite the voltage as
[tex]v(t)=30sin(250t+\alpha )\\[/tex]
a.the impedance of the circuit is express as
[tex]Z=\sqrt{R^{2} +X_{L} ^{2}}\\[/tex]
Where R=200Ω and XL=wL= 250*0.400=100Ω
Hence
[tex]Z=\sqrt{200^{2} +100^{2}}\\ Z=223.6[/tex]Ω
b. The current amplitude is the ratio of the voltage amplitude to the circuit impedance
[tex]I_{max}=\frac{V_{max}}{Z} \\I_{max}=\frac{30}{223.6}\\ I_{max}=0.134A[/tex]
c. the voltage amplitude across the resistor is
[tex]V_{R}=IR=0.134*200\\ V_{R}=26.8v\\[/tex]
And across the inductor we have
[tex]V_{L}=IX_{L} \\V_{L}= 0.134*100\\V_{L}= 13.4V[/tex]
d. the phase angle is calculated as
[tex]\alpha =tan^{-1}(R/X_{L})\\[/tex]
[tex]\alpha =tan^{-1}(200/100)\\\alpha =63.43\\[/tex].
Since it is an R-L circuit, the voltage lead the current.
e. see the attachment.
