![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
There are two bulbs marked with "6V, 3W" and "6V, 6W" respectively. If they are connected in series in the circuit, so that one of the bulbs can light normally, the voltage applied to both ends of the series circuit is () A. 9VB. 12VC. 18VD. 6V
According to the famous brand information, the rated voltage and rated power of the two bulbs, the rated current and bulb resistance of the two bulbs are: rated current I1 = p1u1 = 3w6v = 0.5A, bulb resistance R1 = u1i1 = 6v0.5a = 12 Ω; rated current I2 = p2u2 = 6w6v = 1a, bulb resistance R2 = u2i2 = 6v1a = 6 Ω; the two bulbs are connected in series to make them
"6V 3W" and "3V 6W" bulbs are connected in series in the circuit. What is the current and maximum voltage?
"6V 3W" and "3V 6W" bulbs are connected in series in the circuit. The current is based on the bulb with the largest rated voltage and the smallest power, because its resistance is the largest and the current is the smallest. That is: 1. The rated current of two bulbs in series: I = P / u = 3 / 6 = 0.5 (a) 6V, 3W bulb resistance: r = u × U / P = 6 × 6 / 3 = 12 (Ω) 3V
A small bulb is marked with 6V 3W. When the voltage applied to the bulb is 3V, what is the current and power passing through the bulb?
When the voltage applied to the small bulb is 0? At this time, the resistance of the bulb is? What is the power?
R = u ^ 2 / P = (6V) ^ 2 / 3W = 12 Ω
I1 = U1 / r = 3V / 12eu = 0.25A;
P=U1I1=3V*0.25A=0.75W
If the bulb marked with "6v-3w" is connected to a 3V circuit, the power consumed by the bulb is (3 / 4) W. if the bulb is connected to a 9V circuit and makes it shine normally, a circuit with a resistance value of (6) ohm should be connected in series, and the power consumed by the resistor is (1.5) W
Why 1.5? I don't quite understand,
From 6v-3w, the bulb resistance is 12 Ω, and the current is 0.5 a when the lamp is normally emitting. In the title, it is said that if a 6 Ω resistor is normally emitting, then the current is 0.5 A, the bulb gets 6V voltage, and the resistor gets 3V voltage, from P = UI = 3V * 0.5A = 1.5W
RELATED INFORMATIONS
- 1. Connect the small bulb marked with "6v-3w" into the 3V circuit, and the power consumed by the bulb is___ W? Connect the small bulb marked with "6v-3w" into the 3V circuit, and the power consumed by the bulb is___ W?
- 2. The "3V 3W" lamp is connected to the 1.5V circuit to calculate the actual power
- 3. The bulb L1 marked with 6v6w and the bulb L2 marked with 6v3w are connected in series with the power supply of 12. (1) calculate the actual power P1 and P2 of L, 1 and L2. (2) calculate the total power of the circuit
- 4. Lamp L1 is marked with "8V 16W" and lamp L2 is marked with "12V 36W". The two lamps are connected in series and connected in the circuit with voltage of U What is the maximum voltage to protect the circuit from damage?
- 5. Two small bulbs L1 and L2 are connected in series, and the voltmeter is connected at both ends of L1 and L2 Is it series connection or hybrid connection? Some people say that the voltmeter is disconnected, the current does not go, and the circuit is parallel connection. Is that right?
- 6. Two identical bulbs are connected in series to a constant voltage. If the filament of L1 is broken, the resistance will be reduced. If L1 is still connected in series with L2 to the original circuit, each lamp will be connected What is the difference between the consumed power and the original power?
- 7. Circuit diagram for measuring electric power of small bulb The circuit diagrams that need single a method, double a method, double V method and single V method cannot be disassembled Thank you very much
- 8. Why does the bulb with high resistance share more voltage in series circuit?
- 9. Connect two small bulbs marked with 6v3w. 6v2w in series to both ends of the circuit, and make one of them work normally
- 10. Connect two small bulbs L1 and L2 marked with 6v3w and 6v9w in series and connect them to 8V power supply to calculate the actual power of L1 and L2
- 11. If two small bulbs marked with "6V, 3W" and "6V, 4W" are connected in series, the maximum power allowed to be achieved by the series circuit is______ W. If these two small bulbs are connected in parallel, the maximum power allowed by the parallel circuit is______ W．
- 12. 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
- 13. Two light bulbs, one switch, one power wire, several switches are closed, only one light bulb is on, switch off, two light bulbs are on, how to draw the circuit diagram
- 14. Give you a switch, 2 dry batteries, 2 small bulbs and a number of wires, design the circuit diagram Give you a switch, two dry batteries, two small bulbs and a number of wires. Please design a circuit. When the switch is closed and on, only one light is on. When the switch is off, both lights are on. Draw a design diagram Design scheme and drawings thank
- 15. At present, there are 28 small bulbs in series, which are used as festival lights in the circuit
- 16. There are four light bulbs, which are respectively marked as: (a) "220 V, 60 W" (b) "220 V, 40 W" (c) "220 V, 15 W" (d) "220 V, 100 W." then: (1) the one with the largest resistance is (?) (2) And connected in the lighting circuit, through the current intensity is the largest (?????????) (3) And connected to the lighting circuit, the brightest is (______ (4) in the lighting circuit, the brightest one is (______ ）
- 17. The small bulb L marked with "6V, 3W" and the resistor R with the rated power of 6W are shown in the figure. When the switch S is closed, the lamp l will light normally, and the current indication is 0.75a 1. The value of resistor R 2. The actual power consumed by the whole circuit 3. Xiaoqiang's solution is as follows According to Ohm's law, the resistance of resistor R = u / I = 6V / 0.75a = 8ohm 2. The actual power consumed by the resistor R is PR = 6W, and the actual power consumed by the whole circuit is p = PR + pl (small L) = 6W + 3W = 9W, 3. Please point out the mistakes in Xiaoqiang's solution,
- 18. One resistor is 10 ohm and the other is 2 ohm. Connect these two resistors to the 12V circuit. What is the power of the first resistor
- 19. The small bulb L marked with "6V & nbsp; & nbsp; 3W" and the resistor R with rated power of 4W are connected according to the circuit shown in the figure. When the switch S is closed, the lamp l will light normally, and the current indication is 0.8A. Calculate: (1) the resistance value of the resistor R; (2) the total power consumed by the small bulb and the resistor at this time
- 20. A small bulb with a resistance of 10 ohm normally works at 3V. Now when it is connected to 9V power supply, how large one should be connected in series in the circuit Resistance can make it work normally? (I can't understand this question. I didn't listen to it in class Why subtract 3V from 9V?