Question

An AC voltage source is connected to a 22.0 Ω resistor. The output voltage of the source is given by the function Δv = (100 V)sin[(50π rad/s)t].

(a) From the equation for the output voltage, what is the maximum voltage across the resistor (in V)? V (b) From the equation for the output voltage, what is the frequency of oscillation of the current through the resistor (in Hz)? Hz (c) What is the rms voltage across the resistor (in V)? V (d) What is the rms current in the resistor (in A)? A (e) What is the maximum current in the resistor (in A)? A (f) What is the average power delivered to the resistor (in W)? W (g) What is the current in the resistor (in A) at the time t = 0.0050 s? A

Answer 1

The AC voltage source is ∆V = 100 * sin (50πt)

Now the resistor having value R= 22 ohms .

a) The maximum voltage across the resistor is

Vmax = 100 V ( is as same as the maximum voltage of source voltage )

b)now the frequency of oscillation of current through the resistor is f

w = 50π

2*π*f = 50π

f= 25 Hz

c) Rms volatge across the resistor is Vrms

Vrms = Vmax / √2

V Rms = 100/√2

Vrms = 0.707*100

Vrms = 70.7 V

D) Rms current through the resistor is Irms

Irms = Vrms /R

Irms = 70.7/22

Irms = 3.21 A

E) Maximum current in the resistor is Imax

Imax = =√2* Irms

Imax = √2*3.21

Imax = 4.5 A