摘要：[2005南通]在图16所示电路中.灯泡L标有“6V 3W 字样.滑动变阻器最大阻值R1＝12Ω.先闭合S1.将滑动变阻器滑片P滑至左端.此时灯泡L正常发光．问: (1)电源电压是多大?灯泡L正常发光时的电阻是多大? (2)再闭合S2.电流表示数为0.8A.此时R2的电功率是多大? (3)断开S2.当滑片P滑至右端时.灯泡L的实际功率是多大? [解](1)由于灯L正常发光.所以U = U额 = 6V 灯泡电阻 RL = U额2/P额 = (6V)2/3W =12Ω (2) S1.S2均闭合.R2与灯L并联.I = 0.8A 灯泡中的电流 IL = P额/U额= 3W/6V = 0.5A R2中的电流 I2 = I -IL = 0.8A- 0.5A = 0.3A R2的电功率P2 = U2I2 = UI2 = 6V×0.3A =1.8W (3)P滑到右端时.R1与 L 串联 电路总电流 I 3= U/(R1+RL) = 6V/ = 0.25A 灯泡实际功率P 实 = I32RL = 2×12Ω = 0.75W

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