1. (广元中考)张强妈妈买了一个新的电饭煲,张强从说明书中得知下表中的信息,其工作电路图如图甲所示,$S_{1}$为温控开关,加热电阻$R_{1}和R_{2}$的阻值不随温度变化。求:
(1)电饭煲在“保温”状态下正常工作时,通过电饭煲的电流;
(2)电阻$R_{2}$的阻值;
(3)某周末的晚饭时,张强想利用自家电能表(如图乙)测量家庭电路的实际电压。于是他关闭了家中其它所有用电器,只让电饭煲在“蒸煮”状态下工作,观察到电能表的转盘在1min内转了50转。求家庭电路的实际电压。

解: (1) $ I = \frac{P_{保温}}{U} = \frac{88\ \text{W}}{220\ \text{V}} = 0.4\ \text{A} $;
(2) $ R = \frac{U}{I} = \frac{220\ \text{V}}{0.4\ \text{A}} = 550\ \Omega $,
$ R_1 = \frac{U^2}{P_{蒸煮}} = \frac{(220\ \text{V})^2}{1210\ \text{W}} = 40\ \Omega $,
$ R_2 = R - R_1 = 550\ \Omega - 40\ \Omega = 510\ \Omega $;
(3) $ W = \frac{50}{3000}\ \text{kW} \cdot \text{h} = \frac{50}{3000} \times 3.6 \times 10^6\ \text{J} = 6 \times 10^4\ \text{J} $,
$ P_{实} = \frac{W}{t} = \frac{6 \times 10^4\ \text{J}}{60\ \text{s}} = 1000\ \text{W} $,
$ U_{实} = \sqrt{P_{实} R_1} = \sqrt{1000\ \text{W} \times 40\ \Omega} = 200\ \text{V} $。

2. (毕节中考)如图甲是某茶具上煮茶器的电路原理图,$R_{1}$是加热电阻,$R_{2}$是保温时的分压电阻,$S$为电源开关,$S_{1}$为自动温控开关。$S$、$S_{1}$都闭合时,煮茶器处于加热状态;当水沸腾后,$S_{1}$会自动断开,转为保温状态。煮茶器工作过程中的$P - t$图像如图乙所示,不计电阻值随温度的变化。求:
(1)电阻$R_{1}$的阻值;
(2)当煮茶器处于保温状态时,$R_{1}$的电功率;
(3)当煮茶器处于保温状态时,电阻$R_{2}$在100s内消耗的电能。

解: (1) $ P_{加} = 1100\ \text{W}, R_1 = \frac{U^2}{P_{加}} = \frac{(220\ \text{V})^2}{1100\ \text{W}} = 44\ \Omega $;
(2) $ P_{保} = 44\ \text{W}, I_{保} = \frac{P_{保}}{U} = \frac{44\ \text{W}}{220\ \text{V}} = 0.2\ \text{A} $,
$ I_1 = I_2 = I_{保} = 0.2\ \text{A} $,
$ P = I_1^2 R_1 = (0.2\ \text{A})^2 \times 44\ \Omega = 1.76\ \text{W} $;
(3) $ U_1 = I_1 R_1 = 0.2\ \text{A} \times 44\ \Omega = 8.8\ \text{V} $,
$ U_2 = U - U_1 = 220\ \text{V} - 8.8\ \text{V} = 211.2\ \text{V} $,
$ W_2 = U_2 I_2 t = 211.2\ \text{V} \times 0.2\ \text{A} \times 100\ \text{s} = 4224\ \text{J} $。