88-2 Superposition, source transformation and Thevenin's theorem Superposition 5(-j0 10-i5=-0/A0/s 10(-j5) =2+j4 -10 5-j10 10+j5 104 1=V1+V1"=2-2-1=1-j2 2+4 0.5A R V2=2+V2=j2-2+j2=-2+i4 -110 V1=/x(4-j2)(-j10+2+j4)-4-j28 =2-j2 1/04 4-j2-j10+2+j46-j8 2+14 2 1x(4-212+4)= 4-j2-10+2+j4 -110 2+j4 1=-(-0.5) (4-j2)=-1 4-2 2+i4 105V2=-(-0.5) (4-j2-j10)=-2+j2 6-j8
§8-2 Superposition, source transformation and Thevenin’s theorem Superposition: 4 2 5 10 5( 10) j j j = − − − 10 10 5 10( 5) j j j j j = − − − 2 4 10 5 10( 5) j j j = + + 2 2 6 8 4 28 4 2 10 2 4 (4 2)( 10 2 4) 1 ' 1 j j j j j j j j j V = − − − − = − − + + − − + + = • (4 2) 1 6 8 2 4 ( 0.5) '' 1 − = − − + = − − • j j j V j 2 4 2 10 2 4 (4 2)(2 4) 1 ' 2 j j j j j j V = − − + + − + = • (4 2 10) 2 2 6 8 2 4 ( 0.5) '' 2 j j j j j V j − − = − + − + = − − • V V V 2 j2 1 1 j2V '' 1 ' 1 = 1 + = − − = − • • • V V V j2 2 j2 2 j4V '' 2 ' 2 = 2 + = − + = − + • • • ' 1 • V ' 2 • V '' 1 • V '' 2 • V
Source transformation -10 F07 2+4 05A 4-2 2-|1V 4-j2+2-n1 4-j2-八10+2+/0.6+/0.3 1=4-j2-I(4-j2)=1-j2 V2=I(2+j4)-(2-j1)=-2+j4
Source transformation: 0.6 0.3 4 2 10 2 4 4 2 2 1 j j j j j j I = + − − + + − + − = • V1 = 4 − j2 − I(4 − j2) = 1− j2V • • V 2 = I(2 + j4) − (2 − j1) = −2 + j4V • • 1 • V 2 • V A 0 1
Thevenin’ s theoren: e=1×(4j2)-j0.5(2+j4) 2+14 6-j3 0.5A ztn=(4-j2)+(2+j4)=6+j2 10241 6-j3 4-12 2+4 0.6+j0.3 05A 6+j2-10 -110 1=(4j2)1-(0.6+j0.3=1-j2 6-3v 6+2 2=(2+j4(0.6+j0.3)+j0.5]=-2+j4
Thevenin’s theorem: j V V o c j j j 6 3 1 (4 2) 0.5(2 4) = − = − − + • Zt h = (4− j2)+ (2+ j4) = 6+ j2 0.6 0.3 6 2 10 6 3 j j j j I = + + − − = • V1 = (4 − j2)[1− (0.6 + j0.3)] = 1− j2V • V 2 = (2 + j4)[(0.6 + j0.3) + j0.5] = −2 + j4V •
