soln1 - EE 212 Assignment 1 Solutions Problem # 1 i....

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
EE 212 – Assignment 1 Solutions Problem # 1 i. Voltage and Current Waveforms for the two circuit elements ii. Time period, angular frequency, peak and RMS values of the current i 2 (t) Time period f = 60 Hz T = 1/f = 1/60 = 0.01667 seconds Angular Frequency ϖ = 2 ×π× f = 2 ×π× 60 = 120 π rad/s = 377 rad/s Peak I max = 10 A RMS I rms = 10/ 2 = 7.07 A
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
iii. Instantaneous currents in the two circuit elements when the instantaneous voltage is 10 volts. When v(t) = 10 sin ωt = 10 Or sin ωt = 1 = sin 90 0 Therefore, ωt = 90 0 At ωt = 90 0 : i 1 = 10 sin 90 0 = 10 A and i 2 = 10 sin (90 0 +90 0 ) = 0 A Alternate Solution: From the Figure, When v (t) = 10 volts, Element 1 : i 1 (t) = 10 A Element 2 : i 2 (t) = 0 A iv. Total current i(t) drawn from the source. From Phasor Diagram: I 1 = 10 cm for 10 V I 2 = 10 cm for 10 V (I 1 +I 2 ) = diagonal = 14.1 cm i.e. 14.1 V, at an angle = 45 0 10 cm (10 V) 10 cm (10 V) 14.1 cm (14.1 V) 45 0
Background image of page 2
Therefore, (I 1 +I 2 ) = 14.1 /45 0 A peak (i1 + i2) = 14.1 sin (ωt + 45
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/25/2010 for the course EE 212 taught by Professor Gander during the Spring '10 term at University of Saskatchewan- Management Area.

Page1 / 4

soln1 - EE 212 Assignment 1 Solutions Problem # 1 i....

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online