Answer:
(a) 83.06Ω
(b) 0.81A
(c) 25.0W
Explanation:
Comparing Δv = 95 sin 275t with Δv = Vmaxsinωt
ω = 275
Inductive reactance Χ = ωL
= 275 × 0.240
= 66 Ω
Capacitive reactance Χ = 1/ωc
= 1/ (275 × 26 x 10^-6)
= 139.86Ω
Impedance Z = [tex]\sqrt{R^{2} + (wL - \frac{1}{wc} )^{2}[/tex]
= [tex]\sqrt{R^{2} + (X_{l} - X_{c} )^{2}[/tex]
= [tex]\sqrt{38^{2} + (66 - 139.86)^{2}[/tex]
= 83.06Ω
(b) [tex]I_{rms} = \frac{V_{rms}}{Z}[/tex]
solving for [tex]V_{rms}[/tex],
[tex]V_{rms} = \frac{V_{max}}{\sqrt{2}}[/tex]
[tex]V_{rms} = \frac{95}{\sqrt{2}}[/tex]
[tex]V_{rms} = 67.2V[/tex]
substituting the value of [tex]V_{rms}[/tex] and Z into [tex]I_{rms}[/tex] equation, we have;
[tex]I_{rms} = \frac{67.2}{83.06}[/tex]
[tex]I_{rms} = 0.81A[/tex]
(c) Average power P = [tex]I_{rms}[\tex][tex]V_{rms}[/tex]cos∅
To get the average power, we first solve for ∅ since it was not given.
∅ [tex]= tan^{-1}\frac{X_{l} - X_{c}}{R}[/tex]
∅ [tex]= tan^{-1}\frac{66 - 139.86}{38}[/tex]
∅ [tex]= tan^{-1}\frac{-73.86}{38}[/tex]
∅ [tex]= tan^{-1} -1.9437[/tex]
∅ = -62.77°
Average power P = 0.81 × 67.2 × cos-62.77
P = 0.81 × 67.2 × 0.46
P = 25.03872W
P = 25.0W
