[electricity] Why is the ammeter connected in parallel with a small resistor, and the voltmeter connected in series with a large resistor and a sensitive galvanometer, so the sensitive galvanometer becomes an ideal state [electricity] does sensitive galvanometer have resistance? Why is the ammeter connected in parallel with a small resistance, and the voltmeter connected in series with a large resistance and sensitive galvanometer

[electricity] Why is the ammeter connected in parallel with a small resistor, and the voltmeter connected in series with a large resistor and a sensitive galvanometer, so the sensitive galvanometer becomes an ideal state [electricity] does sensitive galvanometer have resistance? Why is the ammeter connected in parallel with a small resistance, and the voltmeter connected in series with a large resistance and sensitive galvanometer

Sensitive galvanometer with resistance, about several K
Because the ammeter is connected in series in the circuit, the current passing through is large, so a small resistance in parallel with the meter head can make the meter head produce obvious deflection; while the voltmeter is connected in parallel in the circuit, the meter head should be connected in series with a large resistance to limit the current passing through the meter head

How to measure electric power with ammeter and voltmeter?
How to measure electric power with ammeter and voltmeter?

In a circuit, connect a small bulb, the ammeter and the small bulb are connected in series, the voltmeter and the small bulb are connected in parallel, close the switch, read out the voltage and current at this time, and get the power of the small bulb from the formula P = UI

When the range of voltmeter is too small, use sliding rheostat and ammeter, power supply, wire, switch and fixed value resistance to measure the electric power of resistance
Shandong Education Press ninth grade volume II physics "with you to learn new curriculum unit exercise" unit 15 electric power comprehensive test question 27,

Don't draw a very simple series circuit: connect the fixed point of power supply, switch, ammeter, varistor, a certain point, fixed-point lead, bulb, reverse power from another rheostat, the rheostat voltmeter to two points (a fixed point and a fixed point directly connected), the rheostat with the largest resistance, such a circuit resistance, turn on the switch, The voltage is expressed as 6-3.8 = 2.2V,
Please pay attention to the reading of the ammeter when the bulb lights normally
The rated power of the small bulb expressed in P = u and xi = 3.8i

The range of ammeter is 0.6A, the range of voltmeter is 15V, the supply voltage is 36V, R1 is constant resistance, R2 is sliding rheostat
When the resistance of R2 connected to the circuit is 24 Ω, the indication of the ammeter is 0.5A. Now adjust R2 to change the current passing through R1, but ensure that the ammeter does not exceed its range. Question: 1) what is the resistance value of R1? 2) what is the minimum resistance value of R2 connected to the circuit? 3) when R2 takes the minimum value, what is the reading value of the voltmeter

The voltage of R2 is 24 * 0.5 = 12V
The voltage of R1 is 36-12 = 24 v
That is, R1 resistance is 48 ohm
2) It is assumed that the maximum current is 0.6A
The partial pressure of R1 is 0.6 * 48 = 28.8v
R2 partial pressure is 36-28.8v = 7.2V
That is, R2 resistance is 7.2 / 0.6 = 12 Ω
3) The voltage is R2, and the partial voltage is 36-28.8v = 7.2
Absolutely write their own, please give points