a) v(t) = 100 cos(377t +15°) i(t) = 10 sin(377t + 45°) b) v(t) = 100 cos(377t - 75°) i(t) = −10 sin(377t +15°) c) v(t) = √2 100 sin(377t - 30°) i(t) = √2 10 cos(377t – 60°) d) v(t) = 100 sin(377t +30°) i(t) = 10 sin(377t + 120°) e) V = 10 +j10 I=5+j10 f) V = 100 – j90 Ī= 10 - j10 g) V-10-j10 Ī = 5+j10 =

Can you please further explain how you used the ID equation to find the solution? Thank you

Fill in the blanks with the correct choice to complete the sen 1- He always .......... about trivial things. A. Lies B. Lied C. Lying 2- I hope everything is........ Fine tomorrow. A. Will be B. Going to be C. Being 3- She was walking alone When the child............ to cry. A. Begin B. Began C. begun

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Refer to the National Electrical Code® or the working drawings when necessary. Where applicable, responses should be written in complete sentences. For problems 1 and 2, six luminaires, similar to Style E used in the Commercial Building, are to be installed in a room that is 12 ft x 18 ft (~3.7 m x ~5.5 m) with a 9 ft (2.8 m) floor-to-ceiling height. The spacing ratio for the luminaire is 1:0. 1. The maximum distance that the luminaires can be separated and achieve uniform illuminance is ft ( m). For problems 2-5, two luminaires, 8 ft (2.5 m) and 4 ft (1.2 m) in length with dimensions as shown in Figure 15-7, are to be installed in tandem (end to end). The end of the long luminaire is to be 2 ft (600 mm) from the wall. 2. The center of the outlet box should be roughed in at ft ( m) from the wall. 3. The first support should be installed at the wall. ft ( m) from 4. The second support should be installed at from the wall. ft ( m) 5. The final support should be installed at the wall. ft…

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