(a) The maximum voltage V is 190 Volts.
(b) The rms voltage V is 95√2 Volts.
(c) The rms current in Amperes is 7.9 A.
(d) The peak current Amperes is 11.18 A.
(e)The current when t = 0.0045 s is 7.26 A.
The current is the stream of charges which flow inside the conductors when connected across the end of voltage.
Given is an AC source is connected across a 17.0 Ω resistor, the output voltage is given by Δv = (190 V)sin(50πt).
(a) From the given voltage equation, maximum voltage V is 190 Volts.
(b) rms voltage =Vmax/√2
Put the values, we get
Vrms = 190/√2 = 95√2 Volts
(c) rms current = Vrms/Resistance
Put the values, we get
Irms = 95√2 /17
Irms = 7.9 Amperes.
(d) peak current =√2 Irms
Substitute the values, we get
Peak current = 7.9 √2 = 11.18 A
(e) The current when t = 0.0045 s is written as
I = (190 V)sin(50πt)/R
Substitute the values, we have
I = (190 )sin(50π x0.0045)/17
I = 7.26 A.
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