(1) Find V、I(2) Find the absorbing power of each source2Q3A6VSolution:(1) -6+21 +4(I+3)=0I = -1AV = 4(I +3)= 8V(2）Psv =-6I = 6WP3A = -3V = -24W
（1）Find V、I. （2）Find the absorbing power of each source. 6V 4 I 2 3A V Solution： （1） P6V = −6I = 6W P3A = −3V = −24W （2） − 6 + 2I + 4(I + 3) = 0 I = −1A V = 4(I + 3) = 8V
3A22+Find v by superposition method403Q2Solution:O Set the current source to zero@ Set the voltage to zero13A222（+424932324V, =3×(2 // 4) = 4Vx6=4V2+4V = V +V, = 4+4=8V
Find by superposition method. v 6V 3A v 3Ω 2Ω 4Ω 6V v 3Ω 2Ω 4Ω Set the current source to zero 1 4 6 4V 2+4 V = = 3A v 3Ω 2Ω 4Ω Set the voltage to zero 2 V = = 3 (2// 4) 4V 1 2 V V V = + = +4 4=8V Solution：
Simplify the circuit by equivalent transformationOSolution:a626V0ba6VO326V3A9V3A3Q?626 V692boaO32a9929V24V9Vb32Oob
Simplify the circuit by equivalent transformation. 6 3 6V 3A 6V 6 3A 3 a b Solution： 6 1A 3A 6V 3 a b 9V 2 4 V3 6 6V 9V a b a b 9V 9
a2223V(a) Find the Thevenin equivalent circuit12to the left circuit of a-b terminal(b) Find I.306QbSolution:+0+2A2264V3V6 V3 +6Voc+3×642R3S622V3+6ba4266V1.2 A124 +1b
(a) Find the Thevenin equivalent circuit to the left circuit of a-b terminal. (b) Find . I 2A 2 6 I a b 1 3 3V 2 6 a b 3 3 V VOC 2A 4V2 V 0 6 2 2 3 6 V 3 6 V C = + = + = + = + 4 3 6 3 6 R0 2 a b 1 6V4 I 6 1.2 A 4 1 I = = + Solution：