The Student Room Group

Circuits numba 2

circuits.png
V=IR, for the Thermistor, if i increase the temperature the resistance decreases what effect does this have on I through branch 2?

Voltage would decrease because V proportional to R but would I through branch 2 increases slightly or not change and why?

Thanks!
Original post by Zenarthra
circuits.png
V=IR, for the Thermistor, if i increase the temperature the resistance decreases what effect does this have on I through branch 2?

Voltage would decrease because V proportional to R but would I through branch 2 increases slightly or not change and why?

Thanks!


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.
----
eg 1. thermistor resistance Rthermistor= 10kΩ
supply voltage Vs=12V

total resistance Rtotal(thermistor and 10k) = Rthermistor+10k
= 10kΩ+10kΩ
=20kΩ

current I through thermistor (and 10k) =V/Rtotal
=12/20,000
=0.6mA

voltage across thermistor VFD= I*Rthermistor
=0.0006 * 10,000
=6V

voltage across 10k VDB = 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.
-----

e.g. 2 thermistor resistance Rthermistor= 5kΩ
supply voltage Vs=12V

total resistance Rtotal(thermistor and 10k) = Rthermistor+10k
= 5kΩ+10kΩ
= 15kΩ

current I through thermistor (and 10k) =V/Rtotal
=12/15,000
=0.8mA

voltage across thermistor VFD= I*Rthermistor
=0.0008 * 5,000
=4V

voltage across 10k VDB = I*10k
=0.0008 * 10,000
=8V

optional sanity check 8V+4V=12V

----

e.g.3 thermistor resistance Rthermistor=50kΩ
supply voltage Vs=12V

total resistance Rtotal(thermistor and 10k) = Rthermistor+10k
= 50kΩ+10kΩ
= 60kΩ

current I through thermistor (and 10k) =V/Rtotal
=12/60,000
=0.2mA

voltage across thermistor VFD= I*Rthermistor
=0.0002 * 50,000
=10V

voltage across 10k VDB = I*10k
=0.0002 * 10,000
=2V

optional sanity check 10V+2V=12V

---

general result

potential divider.png
Reply 2
Original post by Joinedup
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.
----
eg 1. thermistor resistance Rthermistor= 10kΩ
supply voltage Vs=12V

total resistance Rtotal(thermistor and 10k) = Rthermistor+10k
= 10kΩ+10kΩ
=20kΩ

current I through thermistor (and 10k) =V/Rtotal
=12/20,000
=0.6mA

voltage across thermistor VFD= I*Rthermistor
=0.0006 * 10,000
=6V

voltage across 10k VDB = 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.
-----

e.g. 2 thermistor resistance Rthermistor= 5kΩ
supply voltage Vs=12V

total resistance Rtotal(thermistor and 10k) = Rthermistor+10k
= 5kΩ+10kΩ
= 15kΩ

current I through thermistor (and 10k) =V/Rtotal
=12/15,000
=0.8mA

voltage across thermistor VFD= I*Rthermistor
=0.0008 * 5,000
=4V

voltage across 10k VDB = I*10k
=0.0008 * 10,000
=8V

optional sanity check 8V+4V=12V

----

e.g.3 thermistor resistance Rthermistor=50kΩ
supply voltage Vs=12V

total resistance Rtotal(thermistor and 10k) = Rthermistor+10k
= 50kΩ+10kΩ
= 60kΩ

current I through thermistor (and 10k) =V/Rtotal
=12/60,000
=0.2mA

voltage across thermistor VFD= I*Rthermistor
=0.0002 * 50,000
=10V

voltage across 10k VDB = I*10k
=0.0002 * 10,000
=2V

optional sanity check 10V+2V=12V

---

general result

potential divider.png


Wow, thanks dude for the detailed reply! :smile:

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