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解:在串联电路中,总电压等于各部分电路电压之和,所以$R_1$两端的电压$U_1 = U - U_2 = 8\ \text{V} - 6\ \text{V} = 2\ \text{V}。$
串联电路中电流处处相等,根据欧姆定律$I = \frac{U}{R},$通过$R_1$的电流$I_1 = \frac{U_1}{R_1} = \frac{2\ \text{V}}{4\ \Omega} = 0.5\ \text{A},$此电流也等于通过$R_2$的电流$I_2。$
则$R_2$的阻值$R_2 = \frac{U_2}{I_2} = \frac{6\ \text{V}}{0.5\ \text{A}} = 12\ \Omega。$
答:$R_2$的阻值为$12\ \Omega。$
解:(1)由电路图可知,$R_1$与$R_2$并联,电流表测通过$R_1$的电流$I_1 = 0.3\ \text{A}。$
根据并联电路电压特点,电源电压$U$等于$R_1$两端电压,由欧姆定律$U = I_1R_1$可得:
$U = 0.3\ \text{A} \times 20\ \Omega = 6\ \text{V}$
(2)通过$R_2$的电流$I_2,$由欧姆定律$I_2 = \frac{U}{R_2}$可得:
$I_2 = \frac{6\ \text{V}}{10\ \Omega} = 0.6\ \text{A}$
并联电路总电流$I = I_1 + I_2 = 0.3\ \text{A} + 0.6\ \text{A} = 0.9\ \text{A}$
答:(1)电源电压为$6\ \text{V};$(2)电路的总电流为$0.9\ \text{A}。$

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