# thevenin resistance

Joined Jan 19, 2004
220
was just wondering if thevenin resistances can have negative values...esp in circuits having dependent sources
came across it while solving some sums and even simulated them on multisim...results kinda hold true but still the idea of absolute negative resistance doesnt make any sense..
negative resistance characteristic i hav heard of in UJT, Esaki diodes etc
but i was kinda taken aback with this thevenin resistance being negative

any one who can shine any light on this kindly help
thank you

#### Yeti

Joined Jul 26, 2005
35
I know that capacitors usually are written with negative resistances(impendances). since they are Zc= 1/(jwc) it usually moves the j up top and add a negative. so now you have

Zc = - j/wc.

Thats all the input i have. probably not very helpful though #### Dave

Joined Nov 17, 2003
6,970
Originally posted by Yeti@Sep 25 2005, 08:50 PM
I know that capacitors usually are written with negative resistances(impendances). since they are Zc= 1/(jwc) it usually moves the j up top and add a negative. so now you have

Zc = - j/wc.

Thats all the input i have. probably not very helpful though [post=10592]Quoted post[/post]​
You will find that negative impedance is quite different from negative resistance since they act a right-angles to each other.

Just to make a comment on the opening post. The idea of negative resistance is strange because essentially resistance is a scalar quanity, i.e. a device has a resistance of a particular value (at a particular voltage) independant of the way a voltage is applied across the device. Resistance can be vectorised if we consider the application of a voltage and consider how the current flows in the device taken relative to the Thevenin network, however in reality it is still just a scalar value.

Mathematically speaking:

-Rth = |Rth|

Joined Jan 19, 2004
220
ok consider this circuit
analyse it by two methods is thevenin and say simple ohms/kirchhoff's law
and simulate it on pspice /multisim etc...
the result of thevenin tallies with the ohms law and the multisim results if we take Rth as negative

kindly help

#### aibelectronics

Joined Aug 26, 2005
24
negative resistance is meaningless...
What part of the cct are you isolating? i.e across what node are you doing the thevenin analysis?

Joined Jan 19, 2004
220

#### pebe

Joined Oct 11, 2004
626
Originally posted by haditya@Sep 27 2005, 06:51 AM
ok consider this circuit
analyse it by two methods is thevenin and say simple ohms/kirchhoff's law
and simulate it on pspice /multisim etc...
the result of thevenin tallies with the ohms law and the multisim results if we take Rth as negative

kindly help
[post=10637]Quoted post[/post]​
I'm just curious- what's the symbol on the left hand side of your drawing supposed to be?

Joined Jan 19, 2004
220
a dependent voltage source..in this case a voltage controlled voltage source

#### Dave

Joined Nov 17, 2003
6,970
Ok I'm a little confused so please bare with me.

Am I correct in assuming that the 1kΩ is the load resistor? If so, the thevenin resistance is merely 20kΩ (remove the load and short the voltage sources gives 20kΩ). Where the idea of a negative resistance comes into it I'm not sure. Perhaps you could show your working here for a little clarity.

Again if the load is the 1kΩ resistor then the thevenin voltage is 300V. The current through the 1kΩ is the same as the current through the circuit, and by analysing the thevenin circuit we can see that:

I = 300/(20kΩ + 1kΩ)

I = 14.286mA

Joined Jan 19, 2004
220
in the process to find thevenin resistance we cant ingnore the non zero impedance of the dependent voltage source
in order to find it we apply find the short circuit current Isc and then use it to calculate Rth

to find Vth
we hav 200V1+100-V1=0
ie V1=100/199=-502.5 mV approx.

to find Isc
Isc = 100/20000=5mA

therefore thevenin resistance
Rth= -100.5 ohms or +100.5 ohms ..here where the question lies

****************************************************************

now solving the same circuit by using KVL(assuming i clockwise and lower node as ground)
200V1-20000i+100-1000i=0 and
1000i = V1
therefore
V1= -0.5587 V
i= -558.6 mircoA

there is virtually no disputing this method of solving because the most fundamental laws have been used
*****************************************************************

comming back to thevenin
if we use Rth = -100.5 ohms then the current i=Vth/(Rth+R)
i=-502.5mV/(1000-100.5)ohms
= 558.6 micro A

if we assume Rth = +100.5 ohms
i = -502.5/1100.5
= 456.6 micro A
this result contradicts the one obtained using simple KVL analysis

*******************************************************************
A multisim simulation yeilds this

V1= -558.66 mV
i = 558.66microA anticlockwise

same results show up for an equivalent circuit with -100.5 ohms