Answer Happy

3A,
3,2 COS(3T+53,13)A
3,2 COS(3T+53,13)(mA)
2+3,2COS(3T+53,13)(A)
3+3,2COS(3t-36,87)(A)
- 3-3,2cos( 3t+36,87)(A)
3V,
3,2COS(3t+53,13)(V),
2+3,2cos(3t+53,13)(V),
3,2COS(3t+53,13)(mV)
3 +3 ,2COS (3t-36,87)(V )
-3.3.2cos(3t+36.87)(V)
Using the passive sign convention, determine the voltage defined
as Va:
8 [V]
3,2 Cos (3t + 53,13) [V]
3,2 Cos (3t + 53,13) [mV]
[8 +3,2 Cos (3t + 53,13)] [ V]
[4+3,2 Cos(31-36, 87) ] [V]
[12+ 3,2 Cos(31-127)] [V]
4 [V]
3,2 Cos (3t + 53,13) [V]
6,6 Cos (3t - 22,83) [V]
[-4+6,6 Cos (3t + 22,83) ] [ V ]
[4+3,2 Cos (31-36,8) ) [v]
[12+ 3,2 coseno (3t-127)] [V]
+ Va 4 Ω ΛW - Vb + 2Ω 12 [V] 1 Η 8 coς 3t [V]