Respuesta :
Answer:
Current through resistor [tex]i_{r}=2.5cos(60t+20)mA[/tex]
Current through capacitor [tex]i_{c}=-0.3sin(60t+20)A[/tex]
Explanation:
Given data
Voltage v(t) = 100 cos(60t + 20°) V
Resistance R=40-kΩ
Capacitor C=50-μF
To find
Steady-state currents through the resistor and the capacitor.
Solution
The current flows through the resistor is:
[tex]i_{r}=\frac{v(t)}{R}\\ i_{r}=\frac{100cos(60t+20)}{40*10^{3} }\\ i_{r}=2.5cos(60t+20)mA[/tex]
Now the current flows through capacitor is:
[tex]i_{c}=C\frac{dv(t)}{dt}\\ i_{c}=50*10^{-6}(\frac{d(100cos(60t+20))}{dt})\\ i_{c}=50*10^{-6}*100(-60)sin(60t+20)\\ i_{c}=-0.3sin(60t+20)A[/tex]
The steady state current across the resistor is [tex]i(t) = 2.5 cos (60t+20^\circ) \;\rm mA[/tex].
The steady state current across the capacitor is [tex]i(t) = -0.3 sine (60t+20^\circ) \;\rm A[/tex].
Given that, a voltage v(t) = 100 cos(60t + 20°) V is applied to a parallel combination of a 40-kΩ resistor and a 50-μF capacitor.
The steady state current for the resistor can be calculated by the formula given below.
[tex]i(t) = \dfrac{v(t)}{R}[/tex]
Where, [tex]i(t)[/tex] is steady state current, [tex]v(t)[/tex] is voltage and [tex]R[/tex] is resistor.
Substituting the values in the above formula,
[tex]i(t) = \dfrac {100 cos (60t+20^\circ)}{40\times10^3}[/tex]
Simplifying the above equation as,
[tex]i(t) = 2.5 cos (60t+20^\circ) \;\rm mA[/tex]
So the steady state current across the resistor is [tex]i(t) = 2.5 cos (60t+20^\circ) \;\rm A[/tex].
The steady state current for the capacitor can be calculated by the formula given below.
[tex]i(t) = C\frac{dv(t)}{dt}[/tex]
Where, [tex]i(t)[/tex] is steady state current, [tex]v(t)[/tex] is voltage and [tex]C[/tex] is capacitor.
Substituting the values in the above formula,
[tex]i(t) = 50\times10^{(-6)}\frac{d(100 cos(60t+20^\circ))}{dt}[/tex]
[tex]i(t) = 50\times10^{(-6)}\times100\times(-60)sin(60t+20^\circ)[/tex]
[tex]i(t) = -30\times10^{(-2)} sin(60t+20^\circ)[/tex]
[tex]i(t) = -0.3 sine (60t+20^\circ) \;\rm A[/tex]
So the steady state current across the capacitor is [tex]i(t) = -0.3 sine (60t+20^\circ) \;\rm A[/tex].
For more details, follow the link given below.
https://brainly.com/question/15735611.
