thevenin resistance

Discussion in 'Homework Help' started by haditya, Sep 25, 2005.

  1. haditya

    Thread Starter Senior Member

    Jan 19, 2004
    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
  2. Yeti

    Active Member

    Jul 26, 2005
    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 :(
  3. Dave

    Retired Moderator

    Nov 17, 2003
    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|
  4. haditya

    Thread Starter Senior Member

    Jan 19, 2004
    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
  5. aibelectronics


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

    Thread Starter Senior Member

    Jan 19, 2004
    find current thru the 1kohm
  7. pebe

    AAC Fanatic!

    Oct 11, 2004
    I'm just curious- what's the symbol on the left hand side of your drawing supposed to be?
  8. haditya

    Thread Starter Senior Member

    Jan 19, 2004
    a dependent voltage this case a voltage controlled voltage source
  9. Dave

    Retired Moderator

    Nov 17, 2003
    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
  10. haditya

    Thread Starter Senior Member

    Jan 19, 2004
    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 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
    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)
    = 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