If the two ends of AB and AC are connected to the same power supply, the ratio of the current passing through BC twice is 0

If the two ends of AB and AC are connected to the same power supply, the ratio of the current passing through BC twice is 0

Let the resistance of AB be r and the supply voltage be v. ① the total resistance of AB is 1 / [1 / R + 1 / 3R] = (3 / 4) r. The total current is I (1) = (4 / 3) V / R. the current through BC is I1 = [(4 / 3) V / R] / 4 = (1 / 3) V / R. ② the total resistance of AC is 1 / [1 / 2R + 1 / 2R] = R. the total current is I (2) = the current through BC

The simple circuit diagram of an electric blanket is shown in the figure. R1 is the heating wire of the electric blanket, R2 is the constant resistance. When the switch S is closed, the electric blanket is in the high temperature heating state, and the rated heating power is 110W; when the switch S is open, the electric blanket is in the heat preservation heating state (1) The resistance value of R1; (2) the electric energy consumed in 10min when the electric blanket is heated at high temperature; (3) given that the resistance value of R2 is 660 Ω, what is the actual power of R1 when the electric blanket is heated at low temperature?

(1) R1 = u2p value = (220V) 2110w = 440 Ω; (2) the electric energy consumed in 10min when the electric blanket is in high temperature heating state: w = Pt = 110W × 10 × 60s = 6.6 × 104j; (3) when the electric blanket is in low temperature heating state, R1 and R2 are connected in series, and the current in the circuit: I = ur1 + R2 = 220v440 Ω + 660 Ω = 0.2A; P = i2r1 = (0.2A) 2 × 440 Ω = 17.6w When the blanket is heated at high temperature, the power consumed in 10 minutes is 6.6 × 104j; (3) the resistance value of R2 is 660 Ω, and the actual power of R1 is 17.6w when the blanket is heated at low temperature

Each lamp is marked with "220V 100W" and "220V 40W", and their resistance ratio R1: R2 = 484:1210; connect them in series into the circuit,
Then the voltage ratio U1: U2 = uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu.

The voltage at both ends of the series circuit is proportional to the resistance, so the answer is 484:1210
A little more detail
Two bulbs in series, the current through them is equal, set to I
U1：U2＝IR1：IR2=R1：R2=484：1210 =2:5

If R1 and R2 are connected in parallel in the circuit of u = 110V, the resistance of R1 is 100 Ω, and the total power consumed by R1 and R2 is 300W, then R2 =?

P1＝U^2/R1=110*110/100=121W
P2 = P total - P1 = 300-121 = 179w
P2=U^2/R2 R2=U^2/P2=110*110/179=67.6Ω