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How to calculate the parallel resistance of an ammeter with a range of 100ua and an internal resistance of 1K Ω instead of 1.1ma
100uA=0.1mA=0.0001A
According to the meaning of the title:
1-0.1 = 1 (MA) = 0.001 (a)
The voltage at both ends of the resistor is u = IR = 0.0001 × 1000 = 0.1 (V)
Parallel resistance R = u / I = 0.1 / 0.001 = 100 (Ω)
So the parallel resistance is 100 Ω
A DC ammeter with a full scale deflection current of 50, a resistor as the meter head and a shunt resistor in parallel is made into a DC ammeter with a measuring range of 10mA. The shunt resistor should be ()
A 20
B 20.05
C 10.05
D 10
A DC ammeter with a full-scale deflection current of 50ua, resistance Rg of 2K Ω and shunt resistance R1 in parallel is made into a DC ammeter with a range of 10mA. The shunt resistance R1 should be ().
A 20
B 20.05
C 10.05
D 10 handling questions
10mA = 10000ua, because it is in parallel, so if R1 * (10000-50) = 2000 * 50, R1 = 10.05
In a closed circuit, the voltage of the power supply is a fixed value, and the current I (a) is inversely proportional to the resistance R (Ω)
When the resistance is 4 Ω, the current I should be____ .
2*3=4*R R=1.5
What are the factors that affect the resistance of conductor
The resistance is related to the material, temperature, length and cross-sectional area. When the length and material of the conductor are the same, the larger the cross-sectional area is, the smaller the resistance is. When the cross-sectional area and material of the conductor are the same, the longer the length is, the greater the resistance is. When the cross-sectional area and length of the conductor are the same, the resistance is different with different materials
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