解:(1) 当S 1 闭合,S 2 断开时R 1 被短路,R 3 断路U 源 = IR 2 =0.3A×20Ω=6V;(
| 解: (1) 当S 1 闭合,S 2 断开时R 1 被短路,R 3 断路U 源 = IR 2 =0.3A×20Ω=6V; (2)当S 1 与S 2 均断开时,R 3 断漏配桐卖好路,R 1 与R 2 串联R=R 1 +R 2 =10Ω+20Ω=30Ω I A =U 源 /R=6V/30Ω=0.2A U 1 =I A R 1 =0.2A×10Ω=2V (3)S 1 与S 2 均闭合返坦时,R 1 被短路,R 2 与R 3 并联 I 2 =U 2 /R 2 =U 源 /R 2 =6V/20Ω=0.3A I 3 =U 3 /R 3 =U 源 /R 3 =6V/30Ω=0.2A |