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用十字相乘法解一元二次方程在物理中有不少例子.请看: 例1 如图1所示,额定功率为2瓦的小灯泡与一个阻值为4欧的电阻R串联后接到电压为6伏的电源上,小灯泡恰能正常发光.要使电路消耗的电功率较小,小灯泡的电阻和额定电压的值应为( ) (A)2欧、2伏. (B)2欧、4伏. (C)8欧、2伏. (D)8欧、4伏. (2000年福建) 解析电路消耗的功率是小灯泡的功率和电阻的功率之和,由于小灯泡正常发光,消耗的功率是确定的,因此要使电路消耗的电功率较小,就应使电阻消耗的电功率较小,电阻的功率为
Using quadratic multiplication to solve a quadratic equation has many examples in physics. Look at: Example 1 As shown in Figure 1, a small bulb rated at 2 watts is connected in series with a resistor R of 4 ohms. On a power supply with a voltage of 6 volts, a small light bulb can properly emit light. To make the circuit consume less electric power, the value of the light bulb’s resistance and rated voltage should be () (A) 2 ohms, 2 volts. (B) 2 ohms, 4 volts. (C) 8 ohms, 2 volts. (D) 8 ohms, 4 volts. (Fujian, 2000) The power consumed by the resolver circuit is the sum of the power of the small light bulb and the power of the resistor, since the light bulb is normal. Luminescence, the power consumption is determined, so to make the circuit consumes less electric power, the electric power consumed by the resistor should be made smaller, and the power of the resistor is