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If there is an ammeter with an internal resistance of 0.15 ohm and a range of 1a, then in the same time, if it is paralleled with a resistance of 0.05 ohm, then the range of the ammeter is 1a I hope I can be more detailed
The calculation process is as follows:
First, calculate the voltage when the ammeter is full biased
U=IR=1A*0.15Ω=0.15V
Calculate the total current of the parallel circuit, which is the new range of the ammeter
I1=1A
I2=0.15V/0.05Ω=3A
I=1A+3A=4A
So the range of the new ammeter is 4a
An ammeter with a range of 500 mA and an internal resistance of 100 ohm is refitted into an ammeter with a range of 1a. Then it is connected to a circuit in parallel with a 10 ohm resistor. If the ammeter shows 0.8 a, the current in the circuit is 1 A
A 0.96A
B 1.8 A
C 4.8 A
D 11.2A
C. To change 4.8A into 1A ammeter, a 100 ohm resistor needs to be connected in parallel with the meter head. When indicating 0.8A, the meter head flows through 0.4A, the parallel 100 ohm resistor flows through 0.4A, and the parallel 10 ohm resistor is 1 / 10 of 100 ohm, so the current on the 10 ohm resistor is 10 times of the current on the 100 ohm resistor, which is 4a, and the total current is 0.4 + 0.4 + 4 = 4,8a
What is the range of an ammeter with a maximum range of 0.5A and an internal resistance of 0.15 Ω in parallel with a resistance of 0.05 Ω?
0.5*0.15/0.05=1.5
1.5+0.5=2
An ammeter with an internal resistance of 100 ohm is allowed to pass a maximum current of 500 microampere. How can it be modified into an ammeter with a range of 2.5A
An ammeter with an internal resistance of 100 ohm can pass a maximum current of 500 microampere. How can it be modified into an ammeter with a range of 2.5A
Probably: add resistance to the external circuit, and then change the measuring range
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