The ammeter with a range of 1A has a resistance of 0.1 ohm. What kind of resistance does it need to be refitted into a 6A ammeter? After refitting, the current of a circuit is 0.6A What is the actual current at this time

The ammeter with a range of 1A has a resistance of 0.1 ohm. What kind of resistance does it need to be refitted into a 6A ammeter? After refitting, the current of a circuit is 0.6A What is the actual current at this time

The essence of changing the full current of the original ammeter from 1a to 6a is that the current of 6A flows through the modified part of the original ammeter, and the other 5A is shared by the parallel resistance, so the resistance of 0.02 ohm is used in parallel (2) is 0.6ax6 = 3.6a

Refit the ammeter with maximum range of 0.1A and resistance of 0.5 ohm to 0.5 ohm

Six such ammeters are connected in parallel (only one ammeter can display the reading, and the number can be changed to 6 times of the original one)

An ammeter with a measuring range of 5mA is refitted into ohmmeter X1 gear. The internal resistance of the ammeter is 50ohm, and the electromotive force of the battery is 1.5V. After zeroing, the resistance is measured. When the pointer of the ohmmeter points to the 3 / 4 full deviation position, the resistance value of the measured resistance is?

The ammeter is 5mA, the internal power supply is 1.5V, and the center value resistance ro = 1.5v/0.005a = 300 Ω. Because the internal resistance of the meter head is 50 Ω, a 250 ohm resistance (including the zero resistance) should be connected in series
When the needle points to the 3 / 4 position of full bias, the current I = (3 / 4) * 5mA = 3.75ma;
From I = u / (RO + Rx)
Here, u = 1.5V, RO = 300 Ω, I = 3.75ma = 0.00375a,
It can be found that RX = 100 Ω
See my answer to the same question http://zhidao.baidu.com/question/137017040.html

Try to refit an ammeter with internal resistance of 20 Ω and measuring range of 1mA
1. How much resistance should be connected in parallel for an ammeter with a measuring range of 10A?
2. How much resistance should be connected in series for a voltmeter with a range of 100V?

An ammeter with internal resistance of 20 Ω and measuring range of 1mA is refitted into
1. For an ammeter with a measuring range of 10a, the resistance R shall be connected in parallel
I1R1=(I2-I1)R R=0.002Ω
Parallel resistance R = 0.002 Ω
2. The voltmeter with the range of 100V should be connected in series with resistor Rx
RX=U/I1-R1=99980Ω