Question

In the given circuit, I, = 620°. Determine the average power absorbed by the 40-2 resistor. 1. -j20 2 0.51, 40 2 j10 2 The average power absorbed by the 40-2 resistor is W.

Answer 1

Sz02 V2 co.5To By Applying Kchat node v, -6E +ų + V, -Uz =0 320 -320 > v[-jo1+ioo]-{+io os]= 6L° {-joos] -v[ioos] <6L8 -O By Applying kch al- node Vz V2-V - 0.5I0+ 2 =0 v-V2 S20 320 Io = V-V2 =-> -0. 40 V2-V - 0-S4 +o.SV2 ų 0 -S20 - 820 -320 -820 -uf0.05+jo 025] +V{jo.o5+jo 025+0 025]=0 +z[0.025+io-075]-0 -@ from egeations O f® &y soluing using be Crames's Reele Gjo.05 - So.05 %3D O- 025+jo 075 -30.075

-১০০s ১০০s }১০০৭ ০-০us+১০০ =|-००९)(০०६+০०) -(-১००s-3००) = |3 १5 ४16-31250°-( ४o® A = + ८०xi6 |-१.45 Az -১০০s G० -১০০ = |-১০০:)০) - 6)১০০)| A2= ১০५ ১০५s +6oxib°)-৭५b V2= -१:१3 +১%.५० V,= १.2) )+११.46| v १.2){ ११-46 %3D 40 = = 1-48o1११46 A =(48)x40 P = 4-3.8০