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The ammeter is out of order. There is a voltmeter. How to measure the rated power of the bulb
Connect the switch, power supply, bulb, sliding rheostat in series, and connect the voltmeter with the bulb in parallel. Adjust the sliding rheostat to the minimum value and read out the voltmeter. Then adjust the sliding rheostat to the maximum value and read out the voltmeter. When the sliding rheostat resistance is minimum (zero), the voltmeter reading is equivalent to the power supply voltage U1. When the sliding rheostat is maximum, the voltmeter reading is bulb voltage U2, Using u1-u2 as sliding rheostat voltage and (u1-u2) / sliding rheostat resistance R, the current I2 of sliding rheostat at maximum resistance can be obtained. At this time, the bulb current is I2, the voltage is U2, and the bulb power P = U2 * I2
Series circuit: power supply, switch, bulb, ammeter, voltmeter. In the circuit, the bulb is connected in series with the ammeter and in parallel with the voltmeter, but if they are connected reversely, what will happen
1. The ammeter is burnt out and the bulb is not on
2. The voltmeter and ammeter are intact, and the bulb is not on
2
Because the resistance of the voltmeter is infinite, there is no current in the circuit
The ammeter is connected in parallel with the lamp, and then connected in series with the voltmeter,
Although the ammeter is connected in parallel with the lamp, the front-end circuit is also connected in series with a voltmeter, so there is no current on the ammeter and the lamp. The voltmeter shows the output voltage of the power supply
But if the voltmeter is connected in series with the lamp, then they are connected in parallel with the ammeter, as the first floor said
The voltmeter is connected in parallel with the bulb, the ammeter is connected in series with the bulb, and the bulb is short circuited. What's the change of voltmeter and current indication?
Voltmeter = 0, ammeter larger (burnable meter)
To explore the rated power of the bulb, we selected the equipment: bulb with rated voltage of 3.8V, ammeter and voltmeter, 3 battery, switch, wire. Can you measure it?
Yes, because the voltage is not the same as the rated voltage, the resistance R of the bulb must be measured first,
When two batteries are connected in series, the voltage on both sides of the bulb is U1 and the current is I, r = U1 / I
P = u × U / R
RELATED INFORMATIONS
- 1. When measuring the electric power of small bulb, considering that the ammeter and voltmeter have certain resistance, should the measured value of electric power of small bulb be larger than the real value
- 2. Xiaohong and Xiaowei have done the experiment of "measuring the electric power of small light bulb". There are two kinds of power supply (voltage is 6V), voltmeter, ammeter and sliding rheostat (a is marked with "50 Ω & nbsp; & nbsp; 2A", B is marked with "20 Ω & nbsp; & nbsp; It is known that the rated voltage of the two small lamps are 2.5 V and 3.5 V respectively, and the rated current is the same. They choose the equipment to carry out the experiment. Xiaohong connects the circuit correctly, connects a small lamp into the circuit, closes the key, moves the slide of the rheostat, and observes that the small lamp lights normally when the slide is near the midpoint. The ammeter shows the following As shown in the figure. ① during the experiment, the rheostat selected by Xiao Hong is______ (choose "a" or "B"), the rated voltage of the lamp she measured is______ V, rated current of______ Please calculate the rated power of Xiaohong and Xiaowei according to the relevant information______ (write the calculation process)
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- 11. In the series circuit, a bulb short circuit or open circuit, how does the voltmeter and ammeter change, and in the parallel circuit, how does the ammeter and the voltmeter change
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- 13. If the light bulb is short circuited, does the voltmeter and ammeter show the number?
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