解答下欧姆定律计算题1 一只小灯泡正常发光时的电流是0.5A,灯丝电阻是5Ω,要测量灯泡两端的电压,电压表的量程误码选用
10个回答

1, 灯泡的工作电压为U = I * R = 0.5 * 5 =2.5 V 量程至少大于2.5V

2, R = U / I = 5 / 0.25 =20欧姆, I1 = U1 / R =6 /20 = 0.3A

3, I= U /R =10/20=0.5A, R1= (U1-U)/I = (12 - 10)/ 0.5 = 4 欧姆.