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Troyanec [42]
3 years ago
6

The resistor used in the procedures has a manufacturer's stated tolerance (percent error) of 5%. Did you results from Data Table

agree with the manufacturer's statement? Explain.
Resistor Measured Resistance
100 99.1
Physics
1 answer:
nata0808 [166]3 years ago
8 0

Answer:

     e% = 0.99%   this value is within the 5% tolerance given by the manufacturer

Explanation:

Modern manufacturing methods establish a tolerance in order to guarantee homogeneous characteristics in their products, in the case of resistors the tolerance or error is given by

          e% = | R_nominal - R_measured | / R_nominal 100

where R_nominal is the one written in the resistance in your barcode, R_measured is the real value read with a multimeter and e% is the tolerance also written in the resistors

let's apply this formula to our case

R_nominal = 10 kΩ = 10000 Ω

R_measured = 100 99 Ω

        e% = | 10000 - 10099.1 | / 10000 100

        e% = 0.99%

this value is within the 5% tolerance given by the manufacturer

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Answer:

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x =  vx*t  

x = (22.98)* ( 1.19 )

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