Question

Please give the correct answer with explanations, thankyou

• Find the power delivered to an element at t = 5 ms if the current entering the positive terminal is i = 5 cos 60nt A, and the voltage is as follows: • (Worked solutions of examples are given in class) (a) v = 2i v (b) v = 3 v (c) v=10+5 [i()dt V (Practice ct v= 10 + 5 V (Practice Problem) Answer: (a) p = 17.27 W absorbed (b) p = 6.722 kW supplied (C) p = 29.7 W absorbed

Answer 1

34 Sms (a) We have to find instantaneous power at 2= zi i= 5 los 6ott Power= Voltage & Current Here power is the instantaneous power ri PCt), 7(+) x i(t) - [ 2 i [ty] *i(t) = 23Ct 32556366014)]2 = 2X256353180°= 117 - 50% (0.3459) [ 4354 = 0.345] = 17.27 As power is positive it is absorbed (6) i=5 los bonit b = 3d - 34 (560560T+) = 3x5x6-6N GOTTt) x60TI =-9001 sin bonit Plt) > 206+) Xilt) . 5 cos Gonty (-900 17 Singont) -- 45000 Sinbo Tit as bonit 2 - 450015 x/2 &nbont costanit ] Multiplying Numerator & Denominator by 2] - - 2250 | Sin 1200t i sin 1207185 =-225 plt-sms) = -2250 TT Sin 12071XS =-2250 I Sin (108) =-6722-622 W 2-6,722 kW As power is negative it is supplied 16-10 Scanned with CamScanner

olts = 10+ sjecode =10**[ 58543(bon]Eee] [Peng slobort) = 10+5[ 5 Sileo) 26 1 = 101 3 Sinfont 10lt=5ms) 2107 5 In bon xs 10+ 0.107 ( 124 1000 olo.107 V ta s Съботе ilt=sms) · swisſot 1.6.01 - 5los 54 1000 L = 2.938.A . sms 5 ms Plt=sms) = 20 (tz sms) x ilta sms) = 10.107 x 2.938 = 2 9.7 W Cabsorbed as power is positive Ics Scanned with CamScanner
wWe note that as the current is entering the postive polarity of voltage the positive power will mean power is absorbed and negative power will mean power is supplied by the source.