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

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

In order to make the current a little larger, the total resistance should be reduced, and a resistance in parallel with R can meet the requirements. If a resistance much smaller than R is paralleled, its parallel value will be smaller than the resistance in parallel, and the current will increase too much. Therefore, a resistance much larger than R should be paralleled to make the total resistance slightly smaller and meet the requirements; So a, B, C are wrong and D is correct

The sliding rheostat is connected in series with the fixed value resistor, the voltmeter is connected in parallel at both ends of the sliding rheostat, and the ammeter is connected in series in the circuit to slide the slide of the sliding rheostat
The indication of voltmeter decreases by 4V, the indication of ammeter increases from 0.1A to 0.3A, and the power supply voltage remains unchanged (unknown). Is the resistance of sliding rheostat increased or decreased? How much w has the power of constant resistance changed

Because the current increases, the resistance decreases. When the voltmeter measures the voltage of the sliding rheostat and decreases 4V, the voltage at both ends of the setting resistor increases 4V. If the voltage at both ends of the first setting resistor is u, then u / 0.1 = (U + 4) / 0.3, u = 2V, so the first power is P1 = UI = 2V * 0.1A = 0.2W, and the second power is p

In the circuit, the sliding rheostat is connected in series with the constant value resistor and the ammeter. The voltmeter measures the voltage of the sliding rheostat. When the switch is closed, the slide moves at two points and the current indication changes: 0.5a-1.5a, and the voltmeter 3v-6v to calculate the power supply voltage

The set resistance is R1 supply voltage U
The power supply voltage can be obtained by solving the equations

In a series circuit, a voltmeter measures the voltage of a sliding rheostat. When the resistance of the sliding rheostat becomes small, how do the ammeter and voltmeter change

The voltage indicates that the number becomes smaller and the current indicates that the number becomes larger because voltage = current * resistance and current = voltage divided by resistance