An ideal voltage source provides a constant voltage at the output terminals regardless of the connected load. The case is not same as the practical voltage source. Today you'll learn:

- The circuit diagram and model of ideal and practical sources
- Equations for ideal and practical models
- Models of different practical sources
- Example calculations

### Circuit diagrams and models of both

The internal resistance of an

**is zero. The figure below displays an ideal voltage source connected to a load resistor.***ideal voltage source*
Since the internal source resistance is zero, we can simply replace the circuit by:

The case is not simple with the practical voltage source. Here the internal resistance source is not zero. A practical source always possesses some internal resistance. That can be either 0.002 â„¦ or 1 or 3 â„¦.

### Equations

The current supplied by the ideal source can simply be calculated from the Ohm's law. That is I = V/R. In this case, the source voltage is always equal to load voltage.

The case with the practical source is not simple, here we have to apply the Kirchhoff's voltage law for calculation. Here V

_{load}= V - V_{Rs}.
or V

Let's assume the current I is flowing through the circuit, now:

I R

The equation above explains the current supplied by the practical source. We can also write the equation as I = V/[

Let's solve an example to understand the behavior.

Example: A 10-ohm load is connected to a 5 V ideal voltage source. Calculate the current supplied by the source. Replace the source by practical source which has an internal resistance of 5 ohms and find the current delivered to load.

Solution:

One can easily understand that terminal voltage and thus the current supplied by real-world source depends on its internal resistance.

_{Rs}= V - V_{load }Let's assume the current I is flowing through the circuit, now:

I R

_{s }= V - V_{load }**I = [V/R**_{s}] - [V_{load}/R_{s}]The equation above explains the current supplied by the practical source. We can also write the equation as I = V/[

**R**+_{s }**R**]_{load}Let's solve an example to understand the behavior.

Example: A 10-ohm load is connected to a 5 V ideal voltage source. Calculate the current supplied by the source. Replace the source by practical source which has an internal resistance of 5 ohms and find the current delivered to load.

Solution:

One can easily understand that terminal voltage and thus the current supplied by real-world source depends on its internal resistance.

### Internal Resistance of practical sources

The internal resistance of any battery depends on its construction, design, and quality. However, a good quality 12 v battery has an internal resistance less than 0.002 â„¦ whereas the 1.5 v watch cell could possess 5-10 â„¦. In worst cases, it could be 20 or even 50 â„¦.

We can summarize the above discussions:

We can summarize the above discussions:

Ideal voltage source | Practical |
---|---|

No internal resistance | Possesses some internal resistance |

Current can be calculated by I = V/R | Current is: I = V/[ Rs + Rload] |

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