Question

如图所示的电路中，灯泡L₁、L₂上分别标有“6V 3W”、“6V 9W”的字样，两灯泡的电阻、电源电压均保持不变。当开关S₁闭合，S₂拔至a时，灯L₁正常发光，电压表的示数为6V。 求：

（1）电源电压为多少？

    （2）定值电阻R的阻值为多少？

（3）当开关S₁闭合，S₂拔至b时，

灯L₂消耗的电功率是多少？

Answer 1

（1）当S₁闭合、S₂接a时，灯L₁与R串联，则

        U =U_R + U₁ = 6V+ 6V= 12V······························································· （2分）

（2）由于灯L₁与R串联，且L₁正常发光，则

I_R=I₁＝==0.5A··································································· （1分）

得：R=错误！未找到引用源。=错误！未找到引用源。=12Ω   （1分）

（3）灯L₂的电阻R₂==＝4Ω················································ （1分）

当S₁闭合、S₂接b时，灯L₂与R串联，则

I_R=I₂＝==A···························································· （1分）

灯L₂消耗的电功率为：

P₂＝＝A×4Ω=2.25W·················································· （1分）