As shown in the figure, the power supply voltage is 6V. A small bulb marked with "3V & nbsp; & nbsp; 3W" is connected in series with a sliding rheostat with a maximum resistance of 9 Ω. On the premise of ensuring the safe operation of the circuit, the maximum power consumed by the circuit and the maximum voltage at both ends of the sliding rheostat are () A. 6 W     4.5 VB. 4.5 W   6 VC. 6 W     6 VD. 9 W     4.5 V

As shown in the figure, the power supply voltage is 6V. A small bulb marked with "3V & nbsp; & nbsp; 3W" is connected in series with a sliding rheostat with a maximum resistance of 9 Ω. On the premise of ensuring the safe operation of the circuit, the maximum power consumed by the circuit and the maximum voltage at both ends of the sliding rheostat are () A. 6 W     4.5 VB. 4.5 W   6 VC. 6 W     6 VD. 9 W     4.5 V

(1) Rated current of bulb: Il = PL, UL = 3w3v = 1a, resistance of bulb: RL = UL, IL = 3v1a = 3 Ω; (2) when the bulb is normally emitting, the current Imax = 1a in the circuit is the maximum, and the maximum power Pmax = uimax = 6V × 1A = 6W; (3) when the resistance of sliding rheostat is the maximum, the voltage at both ends of sliding rheostat is the maximum, and the current in the circuit: I '= URL + r = 6v3 Ω + 9 Ω = 0.5A The maximum voltage at both ends of sliding rheostat: urmax = I ′ r = 0.5A × 9 Ω = 4.5V

The two bulbs are respectively marked with "3V & nbsp; & nbsp; 1W" and "6V & nbsp; & nbsp; 3W". If the two bulbs are connected in series to the circuit, the maximum total voltage can be applied at both ends of the bulb to avoid damaging the bulb______ If two bulbs are connected to the circuit in parallel, the maximum current allowed to pass in the main circuit is______ ．

(1) ∵ P = u2r ∵ the resistance of "3V & nbsp; & nbsp; 1W" lamp R1 = u21p1 = (3V) 21W = 9 Ω, the resistance of "6V & nbsp; & nbsp; 3W" lamp R2 = u22p2 = (6V) 23W = 12 Ω, the current in "3V & nbsp; & nbsp; 1W" lamp I1 = p1u1 = 1w3v = 13a, "3V & nbsp; & nbsp; & nbsp; & nbsp; The current of "1W" lamp is I2 = p2u2 = 3w6v = 0.5A, and the two bulbs are connected in series to the circuit, so the current is I = 13a, the total resistance of the circuit is r = R1 + R2 = 9 Ω + 12 Ω = 21 Ω, and the total voltage at both ends of the circuit is u = IR = 13a × 21 Ω = 7V; (2) the two bulbs are connected in parallel to the circuit, and the voltage at both ends of the circuit is equal, so as not to damage the bulb, the voltage at both ends of the circuit should be equal to it The actual current of "3V & nbsp; & nbsp; 1W" lamp: I1 ′ = u ′ R1 = 3v9 Ω = 13a ≈ 0.33a, the actual current of "6V & nbsp; & nbsp; 3W" lamp: I2 ′ = u ′ R2 = 3v12 Ω = 0.25A, so the total current in the trunk road is I ′ = I1 ′ + I2 ′ = 0.33a + 0.25A = 0.58a