The current in the thermistor and 10k resistor is given by I=V/R for that branch
V is 12 and R is 10k+ the resistance of the thermistor.
if the resistance of the thermistor goes down, current goes up through the thermistor and 10k resistor and is the same in both because they are in series .
which voltage are you asking about? is it the voltage between B and D or the voltage between D and F?
components in series with each other carry the same current but may have different voltages across their terminals depending on their contribution to the total resistance- this is the basis of the potential divider.
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eg 1. thermistor resistance R
thermistor= 10kΩ
supply voltage V
s=12V
total resistance R
total(thermistor and 10k) = R
thermistor+10k
= 10kΩ+10kΩ
=20kΩ
current I through thermistor (and 10k) =V/R
total=12/20,000
=0.6mA
voltage across thermistor V
FD= I*R
thermistor=0.0006 * 10,000
=6V
voltage across 10k V
DB = I*10k
=0.0006 * 10,000
=6V
optional sanity check 6V+6V=12V so we're not missing any volts.
big whoop - we've just shown that identical resistances in series divide the supply voltage into two equal halves.
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e.g. 2 thermistor resistance R
thermistor= 5kΩ
supply voltage V
s=12V
total resistance R
total(thermistor and 10k) = R
thermistor+10k
= 5kΩ+10kΩ
= 15kΩ
current I through thermistor (and 10k) =V/R
total=12/15,000
=0.8mA
voltage across thermistor V
FD= I*R
thermistor=0.0008 * 5,000
=4V
voltage across 10k V
DB = I*10k
=0.0008 * 10,000
=8V
optional sanity check 8V+4V=12V
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e.g.3 thermistor resistance R
thermistor=50kΩ
supply voltage V
s=12V
total resistance R
total(thermistor and 10k) = R
thermistor+10k
= 50kΩ+10kΩ
= 60kΩ
current I through thermistor (and 10k) =V/R
total=12/60,000
=0.2mA
voltage across thermistor V
FD= I*R
thermistor=0.0002 * 50,000
=10V
voltage across 10k V
DB = I*10k
=0.0002 * 10,000
=2V
optional sanity check 10V+2V=12V
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general result