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初中应用物理知识竞赛关于电功率的拓展知识主要有:一、用电器串联所允许加的最大电压几个用电器串联,其电流相等.可根据用电器的铭牌确定各个用电器的额定电流,取各个用电器额定电流中最小的值,乘以串联用电器的总电阻,即得用电器串联所允许加的最大电压.例1(2008年河南省初中物理竞赛试题)灯L_1标有“8V 16W”的字样,灯L_2标有“12V 36W”的字样,两灯串联后接在电压为U的电路中.要保证两灯不损坏,电压U的最大值为().A.8 V B.12 V C.16 V D.20 V解析:L_1的额定电流I_1=P_1/U_1=16W/8V=2A,电阻为4Ω;
Applied physics knowledge competition junior high school knowledge on the expansion of electric power are: First, with the maximum allowable series of electrical appliances plus a few appliances in series, the current is equal.According to the electrical nameplate can be used to determine the rated current of each appliance, take each Example 1 (2008 Henan junior high school physics test questions) lamp L_1 marked “8V 16W, the maximum voltage allowed to increase in series with the use of electrical equipment rated current minimum, multiplied by the total resistance of series electrical appliances. ”, Lamp L_2 marked “ 12V 36W ”the words, two lights connected in series after the voltage U. To ensure that the two lights are not damaged, the maximum voltage U () .A.8 V B.12 V C.16 V D.20 V Resolution: L_1 rated current I_1 = P_1 / U_1 = 16W / 8V = 2A, the resistance is 4Ω;