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NuraliBest ResponseYou've already chosen the best response.0
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
 one year ago

NuraliBest ResponseYou've already chosen the best response.0
if correct please click best response.
 one year ago
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