Respuesta :
Answer:
IV. j1.06, capacitive
Explanation:
V(t) = 15cos( 200πt + 65 )
Irms = 10 A .
current leads by 90 degree.
reactance of the load .
Imax = Irms x √2
= 10 x √2
= 14.14 A
Now, Imax = Vmax / reactance
reactance = Vmax / Imax
= 15 / 14.14
= 1.06 ohm
Since current leads the voltage therefore , it must contain capacitance .
the load is capacitative. .
Given Information:
Voltage = V(t) = 15cos(200πt + 65°)
Current = I(t) = 10 A rms
Required Information:
Reactance = X = ?
Answer:
Reactance = 1.061 < -90° Ω
Reactance = -j1.061 Ω
Capacitive Reactance
Explanation:
The reactance is calculated by
X = V(t)/I(t)
V(t) = 15cos(200πt + 65°)
The voltage in polar form can be written as
V(t) = 15 < 65°
Where 15 is the magnitude and 65 is the phase angle
Convert the voltage to rms value
V(t) = 15/√2 < 65°
The current in polar form can be written as
I(t) = 10 < 65° + 90°
Since it was given that the current leads the voltage by 90°
I(t) = 10 < 155°
So the reactance is
X = V(t)/I(t)
X = (15/√2 < 65°)/(10 < 155°)
X = (15/10*√2 < 65° - 155°)
X = 1.061 < -90° Ω
or in rectangular form
X = -j1.061 Ω
Since the phase angle is negative, therefore, this is capacitive reactance.
