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4.D【解析】根据变压器的原理可知变压器不会使交流电的周期改变,由题图乙可知所用交流电的周期为2×10-2s,可求得所用交流电的频率为50Hz,A项错误;由题图乙可知通过R2的电流的有效值为1A由串并联特点可知通过R3的电流(即通过电流表的电流)为0.5A,故副线圈的电流为I2=1.5A,由欧姆定律可知副线圈两端的电压U2=1×10V=10V,由变压器原、副线圈两端电压比与原、副线圈匝数比的关系可知原线圈两端的电压U171=100V再根据原线圈电路的串联关系可知R1两端的电压为Uv=220-100V=120V,B、C项错误;根据变压器原理可知变压器传输的电功率为P=U2I210×1.5W=15W,D项正确。
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