求过程和答案

设电流源左端为点a，右端为点b，左侧3Ω电阻下端为点c。断开4Ω电阻；电流源变换为24v电压源，左负右正，与6Ω电阻串联；等效电压Ubc =Uba + Uac= [ 24 - ( 24 + 6 )/( 6 + 5 + 1 ) * 6 ] + [ 10 * 3/( 3 + 2 ) ]= 9 + 6= 15 v；等效电阻Rbc = Rba + Rac= 6//( 5 + 1 ) + 2//3= 3 + 6/5= 21/5Ω；接入4Ω电阻；4Ω电流 I = ( Ubc + 8 )/( Rbc + 4 )= ( 15 + 8 )/( 21/5 + 4 )= 115/41 A 。