在如图所示的电路中，E为电源，其电动势E＝9.0 V，内阻可忽略不计；AB为滑动变阻器，其电阻R＝30 Ω；L为一小灯泡，其额定电压U＝6.0 V，额定功率P＝1.8 W；S为开关，开始时滑动变阻器的触头位于B端，现在接通开关S.然后将触头缓慢地向A方滑动，当到达某一位置C处时，小灯泡刚好正常发光，则CB之间的电阻应为 ( )
A．10 Ω
B．20 Ω
C．15 Ω
D．5 Ω
答案：B
本题考查串联电路的特点及额定电压、额定功率与电阻的关系，关键是对额定电压、额定功率概念的理解，明确滑动变阻器哪一段电阻接入了电路．错误原因是把接入电路的一段与BC段混淆．
灯泡的工作电流I＝
＝
A＝0.3 A，
因灯泡与滑动变阻器串联，所以通过变阻器的电流为0.3 A.
此时A、C间电阻R1＝
＝
Ω＝10 Ω，
故CB间电阻R2＝R－R1＝20 Ω，故B选项正确．
| 阅读理解。 | ||||||||||||||||||
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