Internal Impedance of A Source

 Internal Impedance of A Source

Internal Impedance of A Source: 

All electrical energy sources have some internal impedance (or resistance*). Due to this internal impedance the source does not behave ideally. When a voltage source supplies power to a load, its terminal voltage drops. A cell used in a torch has a voltage of 1.5v across its two electrodes when nothing is connected to it. However, when connected to a bulb, its voltage becomes less than 1.5v. Such a reduction in the terminal voltage of cell may be explained as follows.

Following figure shows that a 1.5v cell is connected to a bulb. When we say “cell of 1.5v”, we mean a cell whose open circuit voltage is 1.5v. In the equivalent circuit diagram(following in image). the bulb is replaced by a load resistor R_L  (of say, 0.9Ω), and the cell is replaced by a constant voltage source of 1.5V in series with the internal resistance R_s (of say, 0.1Ω). The total resistance in the circuit is now 0.1 +0.9 = 1.0Ω. Since the net voltage that sends current into the circuit is 1.5v, the current in the circuit is

                                        I = (V/R) = (1.5/1.0) = 1.5A

 

The terminal voltage (the voltage across the terminals AB of the cell is the same as the voltage across the load resistor R_L. Therefore,

 

                                    V_AB = I * R_L = 1.5*0.9 = 1.35V 

The voltage that drops because of the internal resistance is

 

= 1.5 - 1.35 = 0.15V

 

Note that, if the internal resistance (impedance) of the cell were smaller (compared to the load resistance), the voltage drop would also have been smaller than 0.15v.

The internal resistance of a source may be due to one or more of the following reasons:

 

1. The resistance of the electrolyte between the electrodes, in case of a cell.

2. The resistance of the armature winding in case of an alternator or a dc generator.

3. The output impedance of the active device like a transistor in case of an oscillator(or signal generator) and rectification-type dc supply. 

Concept of Voltage Source :

 

Consider an ac source. Let V_s  be its open-circuit voltage (I.e. the voltage which exists across its terminals when nothing is connected). And Z_S be its internal impedance. Let it be connected to a load impedance Z_L whose value can be varied, as shown in the following figure.

Now suppose Z_L  is infinite. It means that the terminals AB of the source are open-circuited. Under this condition, no current flow. The terminal voltage V_T is obviously the same as the emf V_S , Since there is no voltage drop across Z_S. Let us now connect a finite load impedance Z_L , and then go on reducing its value. As we do this the current in the circuit in the circuit goes on increasing. The voltage drop across Z_S also goes on increasing. As a result, the terminal voltage V_T  goes on decreasing.

 

 

For a given value of Z_L , the current in the circuit is given as

 

     I = (V_s)/(Z_s + Z_L)

Therefore, the terminal voltage of the source, which is the same as voltage across the load, is

 

V_T = I *Z_L = (V_s / Z_s + Z_L ) * Z_L

= (V_s / I+Z_s / Z_L )

 

From the above equation, we find that if the ratio Z_s / Z_L  is small compared to unity, the terminal voltage V_T  remains almost the same as the voltage V_s.  Under this condition, the source behaves as a good voltage source. Even if the load impedance changes, the terminal voltage of the source. Even if the load impedance changes, the terminal voltage of the source remains practically constant ( provided ratio Z_s / Z_L  is quite small ). Such a source can then is said to be a “good voltage source”.