题目内容
(2008
,四川省南充)如下图所示电路,电源电压不变,




求:
(1)开关S接b时
(2)
开关S接a时电阻R在30s内所产生的热量.
答案:2.56W;180J
解析:
解析:
解: (1)![]() ![]() 接 b时的电流:![]() ![]() (2) 接a时的电流:![]() ![]() 电源电压不变,根据题意有: 0.5(8 +R)=0.4(16+R)R=24 Ω接 a时R发热:![]() 解析:本题事实上是两个串联电路的计算,接 b时R与灯![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |

练习册系列答案
相关题目