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××牌电烤箱
额定电压220V
额定
功率
高温档1800W
低温档500W
电源频率50Hz
下表为一台电烤箱的铭牌,其内部简化电路如图所示,R1和R2均为电热丝,通过S2的闭合与断开实现高温档和低温档之间的切换.求:
(1)电路中R1的阻值.(保留一位小数)
(2)电烤箱在低温档位正常工作时,电路中的电流.(保留一位小数)
(3)电烤箱在低温档位正常工作时,R1的功率.(保留一位小数)

【答案】分析:电源电压U一定,由P=可知,电阻越大,电功率越小;由电路图可知,开关S2断开时,两电阻串联,此时电烤箱功率较小,处于低温档;当开关S2闭合时,只有电阻R1接入电路,此时电烤箱处于高温挡;
(1)已知电源电压与高温挡的电功率,由电功率的变形公式可以求出电阻R1的阻值.
(2)由电烤箱铭牌找出处于低温档时的电功率,由电功率公式的变形公式可以求出正常工作时的电流.
(3)已知电阻与电流,由电功率公式可以求出电阻R1的功率.
解答:解:(1)由电路图可知,开关S2闭合时,
只有电阻R1接入电路,电烤箱处于高温挡,
∴P=
∴电阻R1==≈26.9Ω;
(2)由电路图可知,开关S2断开时,电烤箱处于低温档,
∵P=UI,
∴电烤箱在低温档位正常工作时,电路中的电流:
I==≈2.3A;
(3)电烤箱在低温档位正常工作时,R1的功率:
P1=I2R1=(2.3A)2×26.9Ω=142.3W.
答:(1)电路中R1的阻值是26.9Ω.
(2)电烤箱在低温档位正常工作时,电路中的电流是2.3A.
(3)电烤箱在低温档位正常工作时,R1的功率是142.3W.
点评:本题考查了求电阻、电流及电功率问题,分析清楚电路结构、熟练应用电功率公式及其变形公式即可正确解题.
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