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$0.9\ \text{A}。$