The answer is (a) the power component (b) the consumption element (c) the consumption element (c) the power of 380 v on the rated current is 380/220*60 = the voltage is excessive or even burned. The power of 110/220*60 = 30w on the 110v voltage will be too small, and the light bulbs will be dark or not bright. (1) u-mounted circuit u-opener circuit u-short-circuit circuit (2), carrying end voltage is the end voltage of the power source u (v); the current is rui(a) rui(a) rup 2(w) 22 pur(x) p = 8w series with their rated power. Together, power is their rated power. 1. 6 (a) r=1+2+5+4+3 = 15 (a) (b) 214181611 r r=(b) 4121 12111 1 r=2 (d) 1 and 2 times, with equivalent electrical blockages of 1 (1/1 + 1/2) = 3 times, 1(2/3 + 3) = 3/11, and 2 times to the left, 1/(3/11 + 1/2) = 22/17 =. (e) 2/32 times and 4 times combined with 4/3 last 2/3 and 4/3 times = 4/9 = (f) 1/1/1/1 + 1/3) = 3/2 1/(1/2 + 1/4) = 4/3 = 4/3 3/3 = 3/3 3/2 3/2 + 2 + 2 + 4/3 = 29/6 = (a) (1) k1 closure, 1/(1/8 + 1/4) = 8/3 8/(8/3) = 3 (a) (2) k2 closure 1/(1/8 + 1/4 + 1/2) = 8/7 8/8(8/7) = 7 (a) = 0ireu = 2 = 618 r0 r0 = 2/9 = (a) (2) 0 rrei = 8 (a) = (a) (3) ir2 = 102 = 78 (w) * ir 220 = (w) = 0432 iii i (a) 0251 i (a)
