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Can a 1 / 4W 100 ohm resistor be replaced by a 2W 100 ohm resistor? I bought the wrong resistor. It's a 2W one. Does it work the same way? Is it just a little bigger and can be used instead?
Buy a 2w100ohm resistor? It's OK, because the current of 2w100ohm resistor is larger than that of 1 / 4w100ohm resistor, so it's safe. It's just relatively more power consumption
The resistance R1 is 5 Ω, and the maximum current allowed to pass is 3a. The resistance R2 is 6 Ω, and the maximum current allowed to pass is 2A. If they are connected in series to the power supply, what is the maximum voltage of the power supply? If they are connected in parallel to the power supply, what is the maximum current?
Choose the minimum current in series, just like a two channel Lane becomes a three channel lane, but the traffic volume is still calculated as two channels
If two parallel currents are added together, it's like building three lanes beside two lanes, and the traffic flow is calculated as five lanes
1. If the series resistance R1 + R2 = 11 Ω, the maximum allowable current is 2a, the ohm is fixed; u = 11 * 2 = 22V
2. For parallel resistance, first calculate the maximum withstand voltage of R1 is 5 * 3 = 15V; then calculate the maximum withstand voltage of R2 is 6 * 2 = 12V to ensure the safe operation of the two resistors. Therefore, the maximum operating voltage is 12V, and the current flowing through R1 is 12 / 5 = 2.4a, and the current flowing through R2 is 12 / 6 = 2A. Since it is parallel, the total current is 2 + 2.4 = 4.4a
When a small motor works, the power supply voltage is required to be 6V. If the dry battery is used as the power supply, the power supply voltage is required to be 6V______ If the lead-acid battery is used as the power supply, it needs to be connected in series______ The sections are connected in series
The voltage of one dry battery is 1.5V, so four dry batteries need to be connected in series. The voltage of one lead-acid battery is 2V, and three batteries need to be connected in series
When a small motor works, the power supply voltage is required to be 6V. If the dry battery is used as the power supply, the power supply voltage is required to be 6V______ If the lead-acid battery is used as the power supply, it needs to be connected in series______ The sections are connected in series
The voltage of one dry battery is 1.5V, so four dry batteries need to be connected in series. The voltage of one lead-acid battery is 2V, and three batteries need to be connected in series
Among the following components, a dry battery B storage battery C motor D switch can be used as both power supply and electrical appliance
B. The battery is a power source when discharging, and an electrical appliance when charging!
When a small motor works, the power supply voltage is required to be 3V. If dry batteries are used in series, at least 2 batteries should be used?
Incorrect! For example, nickel chromium battery 5 is 1.2V, lithium battery 2 is 3V
The working voltage of an electrical appliance is 12V, and its resistance is 30 Ω. Now the power supply voltage is 18V. What is the resistance in series
Series partial pressure:
I1=I2
U1/R1=U2/R2
12/30=(18-12)/R2
R2 = 15 Ω
The resistance of the small bulb is 20 ohm. In normal operation, the voltage at both ends is 12V. The current power supply voltage is 18V. Connect the lamp to the power supply,
How to connect a resistor with what resistance value?
According to the series partial pressure law: (18-12) / 12 = R / 20
R = 10 ohm
There is an electric bell. Its resistance is 10 ohm. In normal operation, the voltage at both ends should be 6V. But the existing power supply voltage at hand is 12V
But the current power supply voltage on hand is 12V. To connect the electric bell to this power supply, how much resistance does it need to connect?
In fact, I know the answer, but I don't know how to write the process
R1 = 10 Ω, U1 = 6V, u = 12V
I1 = U1 / R1 = 6V / 10 Ω = 0.6A
Because I = I1 = I2
So r = u / I = 12V / 0.6A = 20 Ω
R2 = r-r1 = 20 Ω - 10 Ω = 10 Ω
A: it needs to be connected in series with a 10 ohm resistor
The normal working voltage of an electric bell is 12V and the resistance is 20 Ω. What should we do to connect the electric bell to a 36V power supply?
Known:
Ask for:
Write clearly
I = u / r = 12V / 20 Ω = 0.6A
R = u / I = 36V / 0.6A = 60 Ω
R series = R total - r = 60 Ω - 20 Ω = 40 Ω
A constant resistance with a resistance of 50 Ω is connected in series
RELATED INFORMATIONS
- 1. R1 = 20 ohm, the total resistance of the circuit is 12 ohm, and the current indication is 0.3A?
- 2. What is the total resistance when one 0.1 ohm resistor is connected in parallel with the other 100 ohm resistor
- 3. There is a line "rx21-8 100r" on the resistance. What is the actual resistance value?
- 4. Ask for small resistance measurement (such as less than 0.1 Ohm)
- 5. It is known that the resistance of the resistance wire is about 10 ohm. The following equipment are available for measuring the resistance of the resistance wire. The equipment to be selected are________ A. Ammeter (0.6A 0.5ohm) B. Ammeter (3a 0.1 Ohm) C voltmeter (3V 6000 Ω) D voltmeter (15V 30000 Ω) E sliding rheostat (0 -- 5 Ω 1a) F sliding rheostat (0 -- 1000 Ω 0.5A) G battery 6V H key
- 6. What are the colors of 300 ohm resistors
- 7. What's the difference between 040206030805 package in resistor and capacitor?
- 8. Chip resistor package 0805 can replace 0603 package, the same resistance?
- 9. The 220 V circuit is connected in series with a 1 m resistor and a CBB capacitor. What is the voltage? I want to get 12 V? What should I do
- 10. After rectifying the 220 V AC diode, add a 400 V 10 V f capacitor, and connect 8 1K resistors to 38 LED lamps. How many large resistors should be connected to? I'm a layman, and I made a LED lamp. As mentioned above, there are 8 1K resistors. The resistor is very hot. Is it right for me to connect it? How should I connect it?
- 11. If two resistors R1 = 10 ohm and R2 = 15 ohm are connected in series, the ratio of power consumed by the two resistors P1: P2 is______ The maximum current allowed to pass through the two resistors is 1 A and 0.6 a respectively. When they are used in series, the maximum voltage allowed to be applied at both ends is______ V
- 12. Resistance R1 = 4 Ω, R2 = 6 Ω, connect the parallel connection to the power supply, the total power consumed by the power supply is 40W, and the output power of the power supply is 38.4w? Resistance R1 = 4 ohm, R2 = 6 ohm, connected in parallel to the power supply, the total power consumed by the power supply is 40 W, the output power of the power supply is 38.4 W, calculate the electromotive force E and internal resistance R of the power supply
- 13. After the resistor R with a resistance value of 10 Ω is connected in series with the lamp L with a rated voltage of 2.5 V, it is connected to both ends of the 4 V power supply, and the switch S is closed. The electric power of the lamp L is measured to be 0.4 W Find the current through R and the rated power of the bulb
- 14. How to distinguish the live line, zero line, ground line, electrical wiring If there are live wires, zero wires, ground wires and electrical appliances at the wall switch socket, how to measure them? The live wires can be measured with electric pen, but the key is others, especially the incoming wires of electrical appliances It's said that we can use a multimeter to measure the resistance. What should we do,
- 15. According to the principle of safe use of electricity, the fuse, switch and resistance wire of electric kettle are respectively connected to the live wire, ground wire and zero wire?
- 16. The voltage of live line and zero line is 220 V, how much is the voltage of two live lines? Why?
- 17. When a power supply is connected with 8 ohm resistor, the current through the power supply is 0.15 When a power supply is connected with 8 ohm resistance, the current passing through is 0.15 a, and when it is connected with 13 ohm resistance, the current passing through is 0.10 A. calculate the electromotive force and internal resistance of the power supply!
- 18. As shown in the figure, the coil with area s, number of turns n, resistance R and resistance R form a closed loop. The ideal AC voltmeter is connected in parallel at both ends of resistance R. the coil rotates uniformly at an angular velocity ω around the rotation axis perpendicular to the magnetic field in a uniform magnetic field with magnetic induction intensity B. suppose the time when the coil rotates to the position shown in the figure t = 0 A. When t = 0, the coil is in the neutral plane, the current flowing through the resistance R is 0, and the reading of the voltmeter is 0b. Within 1 second, the direction of the current flowing through the resistance R changes ω π times C. after connecting another resistance at both ends of the resistance R, the reading of the voltmeter will decrease D After a capacitor with larger capacitance is paralleled at both ends of the resistance R, the reading of the voltmeter will not change, and the signal processing system will count every time it gets a high voltage
- 19. The magnetic flux through a single turn coil with a resistance of 1 ohm changes: it decreases evenly in T1 time per second, and it decreases evenly in T2 time per second When the magnetic flux through a single turn coil with a resistance of 1 ohm changes: in T1 time, it decreases uniformly every second, and in T2 time, it increases uniformly every second, then the average value of the induced electromotive force generated in the coil must be zero in (T1 + T2) time. Why is it right? Let's not say that T1 = T2
- 20. The electromotive force of the power supply is e = 30V, the internal resistance is r = 1 ohm, and the coil resistance of the DC motor is r = 2 ohm. If the reading of the ideal ammeter is 2a, the voltage at both ends of the motor? The output power of the motor? (motor and er in series)