![](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>)
How to know which electric current is measured by ammeter? I know which consumer the ammeter is connected in series with is to measure which consumer's current, but I just can't tell which consumer the ammeter is connected in series with. I don't know whether it's left or right!
What I said upstairs is a little complicated. The simplest way is to measure the current of an electric appliance, no matter whether the electric appliance is on the left or right side of the ammeter, as long as the wire connecting the electric appliance and the ammeter is not bifurcated
If the ammeter is connected in parallel in three series circuits with 10 ohm resistance and the current is 1.5 V, what is the supply voltage?
The ammeter is connected in parallel with one of them, and the current indication is 1.5 Amperes
How to parallel, and one of the three are, three words should also slide the rheostat, otherwise the power short circuit, and ammeter how is 1.5V, you make the topic clear
From the known closed switch, the resistance in parallel with the ammeter is short circuited, I = 1.5A, r = 10 Ω * 2 = 20 Ω, u = IR = 1.5A * 20 Ω = 30V
It is known that R1 = R2 = 10 ohm, R3 = 20 ohm. Connect them in series and connect them at both ends of the circuit with constant voltage. Xiaofang parallels an ammeter at both ends of R2, as shown in the figure. It is found that the indication number is 1a. If the ammeter is replaced by a voltmeter, the voltage indication number is? R1 R2 R3. The series voltmeter measures R2 voltage!
If the ammeter is connected in parallel at both ends of R2, it is equivalent to short circuiting R2. The measured ground current is the current of R1 and R3 in series in the circuit. Therefore, the power supply voltage can be obtained as u = IR = (10 + 20) X1 = 30V. If the ammeter connected in parallel at both ends of R2 is replaced by a voltmeter, the total resistance in the line will become R1 + R2 + R3 = 40 ohm. The power supply voltage has been calculated as 30V, Then the line current is 30 / 40 = 3 / 4A, and the voltage at both ends of R2 is u = IR = 3 / 4x10 = 7.5V
Connect a 5 ohm resistor and a 15 ohm resistor in series to a 6-volt power supply. What is the current in this series circuit?
If parallel, what is the current on the main line
1. Series connection:
R=R1+R2=5Ω+15Ω=20Ω
I=U/R=6V/20Ω=0.3A
2. Parallel connection:
R'=R1R2/(R1+R2)=5Ω×15Ω/(5Ω+15Ω)=15/4 Ω
I=U/R'=6V/(15/4 Ω)=1.6A.
RELATED INFORMATIONS
- 1. The two lamps are connected in parallel, the known resistance R1 = 5 ohm, the main circuit current I = 0.5A, the voltage at both ends of R2 U2 = 2V, and the circuit current I = 0.5A The two lamps are connected in parallel, the known resistance R1 = 5 ohm, the driving current I = 0.5A, and the voltage at both ends of R2 U2 = 2V 1. Find the current through R1 2. Current through R2 3. Resistance of R2 It's a process. Thank you
- 2. When a fixed value resistor is applied with 10V voltage at both ends, the current is 0.5A. If the voltage at both ends of the resistor is 0, what is the resistance value?
- 3. What is the resistance value of a resistor when a voltage of 10V is applied at its two ends and the current is 0.5A?
- 4. How many ohm resistor should be connected to 120 white LED lamps connected to 12 volt dry battery? The voltage will drop if there are too many
- 5. 1. The fruit store brought in 180 kg of fruit, sold 2 / 5 of the fruit on the first day, and sold the remaining 3 / 4 the next day. How many kg of fruit are left? 2. The school canteen bought a barrel of oil, which weighs 24 kg. The first time it used 1 / 4 of the total oil, the second time it used the remaining 1 / 3, and the third time it used 1 / 2 of the oil in the barrel. How many kg of oil is left in the barrel?
- 6. Very simple score calculation problem Oh, help to calculate, OK One eighth multiplied by two is equal to one Twelve times three eighths is equal Eight out of twenty-five divided by four out of five equals Four fifths divided by three fifths is equal to four fifths Two divided by two thirds is equal to two Two out of eleven times eleven out of four is equal to one Four ninths divided by three fifths is equal to three fifths Five eighths divided by five twelfth equals five Half of a thirtieth divided by a quarter equals half
- 7. Calculation of atmospheric pressure For a pressure cooker with a diameter of 20cm, the cross section of the vent is 10mm ^ 2. If the maximum pressure of steam in the cooker is 1.7 * 10 ^ 5pa, what is the mass of the pressure limiting valve? (set the external pressure as 1 * 10 ^ 5pa)
- 8. 4 Calculation questions Zero - half - (minus three fourths) + (minus five sixths) - two thirds |Negative two and a half | - (- 2.5) + 1 - | - 1-two and a half| 13-[26-（-21）+（-18）] [1.4-（-3.6+5.2）-4.3]-（-1.5） We need a process. Thank you
- 9. Four practical questions on how many points to get
- 10. The correct statement about conductor resistance is () A. Because the resistance of the conductor indicates the resistance of the conductor to the current, when there is no current in the conductor, the resistance of the conductor is zero. B. the greater the voltage at both ends of the conductor, the greater the current, so the resistance of the conductor decreases with the increase of the voltage. C. the resistance of the conductor is the property of the conductor to block the current, and its size has nothing to do with the voltage and current intensity The greater the voltage at both ends of the conductor, the greater the current, so the resistance of the conductor increases with the increase of voltage
- 11. The voltage at both ends of a circuit is 220 V, and the resistance of the consumer is 40 Ω. Can we use an ammeter with a range of 0 ~ 3a to measure its current?
- 12. The resistance of an electrical appliance is 100 ohm, and its maximum allowable current is 100 mA. If it is connected to a circuit with a current of 1.1 A, what method should be adopted?
- 13. The small bulb marked "6V 3W" is connected in series with a sliding resistor with a maximum resistance of 10 ohm in the circuit. When the sliding resistor is connected to the circuit with a resistance of 6 ohm, the lamp will light normally Ask: 1: bulb resistance, power supply voltage? 2: How much heat does the sliding rheostat generate when the bulb lights normally for 1 minute?
- 14. A small bulb marked with "6V. 3W" is connected in series with a sliding rheostat with a maximum resistance of 30 ohm. When the resistance of the rheostat entering the circuit is 12 ohm, the lamp will light normally. When the resistance of the rheostat entering the circuit is 18 ohm, the bulb resistance and power supply voltage are calculated (2) Voltage at both ends of rheostat (3) Actual power of bulb
- 15. There are two bulbs in series in the circuit. The voltage at both ends of bulb L1 is 3V, the total resistance of the circuit is 10 ohm, and the resistance of bulb L1 is 5 ohm (1) Resistance of bulb L2 (2) current in circuit (3) voltage at both ends of bulb L2 (4) supply voltage
- 16. There is a small bulb with a resistance of 10 Ω. When it lights normally, the voltage at both ends is 6V. Now there is only a 9V power supply Q: 1. How to connect a small light bulb to make it shine? 2. How much resistance do you need to give it? There are three ways to answer the second question!
- 17. The sliding rheostat is connected in series with the ammeter and the lamp, and the lamp and the voltmeter are connected in parallel
- 18. Using voltmeter and ammeter to measure the resistance of small bulb when it is normally emitting, what problems should be paid attention to
- 19. A voltmeter with an internal resistance of 1000 ohm has a measuring range of 3V. When a new voltmeter is assembled by connecting a 5000 ohm divider in series, the reading is 2V A voltmeter with an internal resistance of 1000 ohm has a measuring range of 3V. When it is assembled into a new voltmeter with a 5000 ohm partial resistance in series, the reading is 2V. What is the measured voltage? What is the measuring range of the new voltmeter? How much is the measuring range
- 20. In the circuit as shown in the figure, the power supply voltage is 9 V, the range of voltmeter is 0 ~ 3 V, the range of ammeter is 0 ~ 0.6 a, the setting resistance R1 = 20 Ω, and the sliding rheostat is marked with "15 Ω & nbsp; Calculation: ① when the reading of ammeter is 0.4 a, the voltage at both ends of resistance R1. ② the maximum resistance value of sliding rheostat R2 allowed to be connected into the circuit. ③ in order to make each ammeter reach the maximum value respectively in the process of moving slide, the appropriate fixed value resistance R is selected to replace the fixed value resistance R1 to find out the value range of r that meets the requirements