Respuesta :
Answer:
Explanation:
Rate of dissipation of energy at resistor
= I² R = 296
current I² = 296/R = 296 / 300
I = 0.9933 A.
Impedence Z = V / I = 500 / .9933
= 503 ohm.
B ) Let it be V
voltage across resistance = 300 x .9933 = 298 v
Now 500² = 298² + V²
V = 401 v
C) Power factor = R / Z
=300 / 503
= 0.596
Answer:
Part A)
The impedance of the circuit [tex]Z=503ohm[/tex]
part B)
The amplitude of the voltage across the inductor[tex]V=401V[/tex]
Part C)
The power factor [tex]=596[/tex]
Explanation:
Given:
[tex]$\mathrm{R}=300 \Omega$[/tex]
voltage amplitude [tex]500 V[/tex]
Rate at which electrical energy is dissipated [tex]296W[/tex]
A)Rate of dissipation of energy at resistor
[tex]$=\mathrm{I}^{2} \mathrm{R}=296$[/tex]
current [tex]$\mathrm{I}^{2}=296 / \mathrm{R}=296 / 300$[/tex]
[tex]$\mathrm{I}=0.9933 \mathrm{~A}$[/tex]
Impedence [tex]$Z=V / 1=500 / .9933$[/tex]
[tex]$=503 \mathrm{ohm}$[/tex].
B) Let it be [tex]$\mathrm{V}$[/tex]
voltage across resistance [tex]$=300 \times .9933=298 \mathrm{v}$[/tex]
Now [tex]$500^{2}=298^{2}+V^{2}$[/tex]
[tex]$\mathrm{V}=401 \mathrm{v}$[/tex]
C) Power factor [tex]$=\mathrm{R} / \mathrm{Z}$[/tex]
[tex]$=300 / 503$[/tex]
[tex]$=0.596$[/tex]
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