A certain voltage is +11.1 V for 21.8 ms and -11.1V for the succeeding 21.8 ms and continues oscillating back and forth between these two values at 43.6 ms interval The voltage is present across a 59 12 resistor. Over any 43.6 ms interval, find: (a) the maximum value of the voltage (b) the average value of the voltage (c) the average value of the resistor current (d) the maximum value of the absorbed power (e) the average value of the absorbed power. (a) V (b) Vy= (c) Farg (d) Par (e) Parg = A W W
A certain voltage is +11.1 V for 21.8 ms and -11.1V for the succeeding 21.8 ms and continues oscillating back and forth between these two values at 43.6 ms interval The voltage is present across a 59 12 resistor. Over any 43.6 ms interval, find: (a) the maximum value of the voltage (b) the average value of the voltage (c) the average value of the resistor current (d) the maximum value of the absorbed power (e) the average value of the absorbed power. (a) V (b) Vy= (c) Farg (d) Par (e) Parg = A W W
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Transcribed Image Text:A certain voltage is + 11.1V for 21.8 ms and -11.1 V for the succeeding 21.8 ms and continues oscillating back and forth between these two values at 43.6 ms intervals
The voltage is present across a 59 12 resistor.
Over any 43.6 ms interval, find:
(a) the maximum value of the voltage
(b) the average value of the voltage
(c) the average value of the resistor current
(d) the maximum value of the absorbed power
(e) the average value of the absorbed power.
(6) Vmax=
(b) Vary
(c) Kang
(d) Par
(e) Parg
R
V
A
W
W
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VIEWStep 3: Determine the maximum value of the voltage
VIEWStep 4: Determine the average value of the voltage
VIEWStep 5: Determine the average value of the resistor current
VIEWStep 6: Determine the maximum value of the power absorbed
VIEWStep 7: Determine the average value of the absorbed power
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