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  1. Nurali
    • one year ago
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    If the output is to be set at 6.0 V, then across the other section (A), the voltage drop needed to be 3.0 V. At 0730, the resistance of the thermistor = 1.5 kΩ The current flowing through section (A): I=3V/1.5kΩ+1.5kΩ I=0.001 A Similarly, the same current flows through the section (B) 6.0 V =(Rv+1.5k)*(0.001A) Rv=4.5kΩ

  2. Nurali
    • one year ago
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    if correct please click best response.

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