Input Impedance of SIMPLE Circuit

AI Thread Summary
The discussion revolves around calculating the input impedance of a specific circuit and clarifying the definitions of input and output impedance. Output impedance is equated to Thevenin impedance, while input impedance calculations can be complex when voltage/current sources are present. The input impedance for the discussed circuit is determined to be 100 ohms, but real-world applications may differ significantly from theoretical calculations. Factors such as internal resistance of components and the effects of DC voltage on circuit behavior are highlighted as critical considerations. Accurate impedance measurements require careful analysis of voltage and current under varying load conditions.
gayapatel
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Can anyone figure out the input impedance of the circuit I have attached?

Does anyone know a THOROUGH definition of input impedance? output impedance?

Output impedance is equivalent to the thevenin impedance, which mean we turn off all voltage/current sources. But for input impedance, what if the circuit has voltage/current sources, what do we do (like the circuit I have attached)?
 

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Still no answer :-(

Still no answer from PhysicsForums :-( :frown:
 
Damm.. i had a reply , but it disappeared when i clicked on your attachment ..
your first attachment , if you flip it over to the right ,and short the voltage source just like an output impedance ,what have you got??
 
Your second attachment looks ok ..
 
Convert the thevenin equivalent norton equivalent, that is a shunt resistor and a current source equal to the thevenin voltage divided by 100 ohms. The current source is supposedly infinit so doesn't affect the input impedance calculation.
Clearly the input impedance is 100 ohms.
Joe
 
The circuit you have shown can be looked at several ways:

(1) You can ignore the battery and replace it with an equivalent resistor representing it's internal resistance. Now you can calculate the 'impedance', and since there is no reactive component it is simply resistance with no phase shift.

(2) However, you may be misleading yourself as to how the circuit actually behaves. While the theoretical basics are good for simple circuits, real circuits rarely conform, and the results will not be anything like what you learn in class:

For example, suppose (since we are talking about impedance) that this is an audio application. Applying a D.C. voltage backwards into the output leads of an amplifier can have all kinds of disastrous effects on both the performance and the reliability of the circuit. So, although the 'impedance' we calculated is correct, the actual behaviour of the circuit could be bad. Even a small D.C. voltage applied backward might change the bias on a transistor or tube circuit, and result in the component operating in a non-linear part of its performance specs.

Likewise, one rarely applies A.C. to a drycell, so, this could be just a representation of the input circuit of an op-amp or some other useful device. This means that although there is a 'load resistance' and D.C. voltage peeking out of the input leads, there must be more attached to this circuit than actually appears. Applying an A.C. signal for instance may or may not be able to penetrate other hidden components connected to the system, like capacitors or inductors. Obviously, the circuit will not behave as assumed by the diagram.
 
The input impedance is 100 Ohms plus the output impedance of the battery (frequency dependent). The impedance is deltaV/DeltaI, not V/I. If you are interested in the DC Zin, you need to measure the V and I at a couple of load currents, and use the load line to give you the Z. If you are interested in the Z at a particular frequency or range of frequencies, use a signal generator through a series resistor to find deltaV/deltaI.
 
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