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How to connect the resistance and how to calculate the range of a meter with enough sensitivity? I'll give you more points for details
To change to a voltmeter, a larger resistance R should be connected in series. Range calculation: according to the principle of series voltage division, it should be noted that the meter resistance Rg, the maximum meter current Ig and the corresponding range u can be obtained from u = igrg + IGR, r = u / Ig - RG. If the expanded range voltage is n times of the maximum meter voltage, r = (n-1) RG should be changed to current
Why is it that a resistor in series with the meter head is equivalent to a voltmeter, while a resistor in parallel with the meter head is equivalent to an ammeter,
The voltage measurement of a circuit diagram is accurate, and the current includes the current of the voltmeter. The current is too large. When R is far less than the internal resistance of the voltmeter, the current flowing through the voltmeter can be ignored, and the measurement result is more accurate. Considering the large internal resistance of the voltmeter, a is suitable for measuring the smaller resistance
B circuit diagram current measurement is accurate, the voltage includes the Ammeter voltage, the voltage is too large, when R is far greater than the ammeter internal resistance, the Ammeter voltage can be ignored, the measurement results are more accurate, considering the ammeter internal resistance is larger, so B is suitable for measuring larger resistance
RELATED INFORMATIONS
- 1. There is a power supply resistance switch in the circuit. When the voltmeter is in series in the circuit and the ammeter is in parallel in the circuit, the consequence is that both the ammeter and the voltmeter are There is a power resistance switch in the circuit. When the voltmeter is connected in series and the ammeter is connected in parallel, the result is a. the ammeter and the voltmeter are damaged. B. the ammeter is damaged, and the voltage representation is zero. C. The current representation is unchanged, and the voltage representation is reduced. D. the current representation is reduced, and the voltage representation is reduced
- 2. Why to expand the range of DC ammeter need parallel resistance shunt on the meter, why to expand the range of DC voltmeter need series resistance on the meter
- 3. Use voltmeter and ammeter to measure resistance. Why use sliding rheostat in this experiment
- 4. As shown in the figure, during the movement of slide P of rheostat R0, the indication range of voltmeter is 0 ~ 4V, the indication range of ammeter is 1 ~ 0.5A, and the internal resistance of power supply can be ignored. Calculate the resistance value of resistance R, the maximum value of rheostat R0 and the electromotive force E of power supply
- 5. When Ohm's law is used to measure resistance, if the slide of sliding rheostat is moving, the degree of voltmeter will change from 1.2V to 1.6V, while the degree of ammeter will change from 1.2V to 1.6V—— When using Ohm's law to measure resistance, if the degree of voltmeter changes from 1.2V to 1.6V and the indication of ammeter increases by 0.2A, the resistance value of resistance R to be measured is___
- 6. When measuring resistance R by voltammetry, if the reading of ammeter is 0.2A and that of voltmeter is 30V, what is the resistance value of the resistance to be measured?
- 7. The power supply voltage is 6 V, the indication of ammeter A is 2.5 A, R1 = 4 Ω, calculate the indication of ammeter a1.a2 and the resistance of resistance R2 Well
- 8. Han Yong designed the circuit as shown in the figure to measure the resistance value of resistance R1. When he contacts the switch with a and B respectively, the ratio of ammeter indication is 5:2. Given that R2 = 6 Ω, what is the resistance value of R1?
- 9. The voltage at both ends of the power supply is 6V, R1 and R2 are constant resistance. After closing switch s, the indication of ammeter is 0.2A, and that of voltmeter is 4V What are the resistance values of R1 and R2? Write out the process, too```
- 10. As shown in the figure, U1 = 6V, u = 10V, the current passing through R1 is 0.4A, and the resistance of R1 R2 can be calculated The circuit diagram is patchy___________ I ————|_________ |——————|___________ |————＞—— R1 R2 |＜——U1——＞| |————U2——————＞| |＜——————————U————————————＞|
- 11. In the circuit, r = 10 Ω, the power supply voltage remains unchanged. If a resistor with R1 = 30 Ω is connected in series in the circuit, the indication of the ammeter will change to I1 = 0.25A 1. What is the power supply voltage equal to 2. If the indication of I2 is 0.4A, how to connect the circuit in the diagram? (R1 removed) what is the resistance value?
- 12. One voltmeter is composed of ammeter g and resistance R in series, as shown in the figure. If it is found that the reading of this voltmeter is slightly smaller than the accurate value in use, which of the following measures can be adopted for improvement () A. Connect a much smaller resistance in series on R. B. connect a much larger resistance in series on R. C. connect a much smaller resistance in parallel on R. D. connect a much larger resistance in parallel on R
- 13. One voltmeter is composed of ammeter g and resistance R in series, as shown in the figure. If it is found that the reading of this voltmeter is slightly smaller than the accurate value in use, which of the following measures can be adopted for improvement () A. Connect a much smaller resistance in series on R. B. connect a much larger resistance in series on R. C. connect a much smaller resistance in parallel on R. D. connect a much larger resistance in parallel on R
- 14. The resistance ratio of the two resistance wires is 3:2, and the current ratio through them is 1:4. If they generate the same heat, the power on time ratio is 3:2——
- 15. The conductivity of copper conductor is better than that of aluminum conductor, so the resistance of copper conductor is smaller
- 16. As shown in the figure, a parallel metal guide rail with no resistance is placed in the uniform magnetic field. The guide rail is connected with the large coil m, and a wire AB is placed on the guide rail. The magnetic induction line is perpendicular to the plane of the guide rail. In order to make the small closed coil n surrounded by M produce clockwise induced current, the motion of the wire may be () A. Moving to the right at a constant speed B. accelerating to the right C. decelerating to the right D. accelerating to the left
- 17. 11. As shown in figure 11-1-6, the electromotive force of the power supply is 3V, the internal resistance is ignored, the mass of the conductor rod is 60g, the length is 1m, the resistance is 1.5 Ω, the direction of the uniform magnetic field is vertical upward, and the size is 0.4t. When the switch s is closed, the rod slides from the bottom end of the fixed smooth insulating ring to a certain position and stops (1) What is the pressure of the rod on each ring at this position? (2) If the radius of the insulating ring is known to be 0.5m, what is the height difference between this position and the bottom of the ring? E S B Figure 11-1-6
- 18. When the smooth parallel metal guide rails are placed horizontally with a spacing of 0.5m, the electromotive force of the power supply is e = 1.5V, the internal resistance is r = 2.0 ohm, and the resistance of the metal bar is r = 2.8 ohm, which is perpendicular to the parallel guide rails, and the other resistances are not included. When the metal bar is in a uniform magnetic field with magnetic induction intensity B = 2.0T and the direction is 60 ° to the horizontal direction of water, the circuit will be connected, What is the magnitude and direction of the ampere force on the metal bar? If the mass of the metal bar is m = 5 * 10 negative quadratic kg, what is its pressure on the orbit?
- 19. When the distance L of the metal guide rail is fixed horizontally, the resistance is neglected. The left end of the guide rail is connected with a resistance R. the uniform magnetic field B is perpendicular to the plane of the guide rail downward. The conductor bar with mass m, resistance R and length L is vertically placed on the guide rail to keep good contact with the guide rail. The dynamic friction factor between the conductor bar AB and the guide rail is μ (1) The direction and magnitude of the induced current in the conductor bar AB; (2) the electric power of the resistance R; (3) the magnitude of the external force F
- 20. A conducting rod AB is horizontally placed on two metal guide rails and is in a vertical upward uniform magnetic field. As shown in the figure, the mass of the conducting rod is 1.2kg and the length is 1m. When 3A current is applied to the conducting rod, it can slide on the guide rail at a uniform speed. If the current intensity increases to 5A, the conducting rod can obtain an acceleration of 2m / S2, and the magnetic induction intensity of the device is calculated