Consider a practical voltage source shown in the Fig. 1.(a) having internal resistance R

_{se}, connected to the load having resistance R_{L }.Fig. 1(a |

Now we can replace voltage source by equivalent current source.

**Key point**: The tow sources are said to be equivalent, if they supply equal load current to the load, with same load connected across its terminals.

The current delivered in above case by voltage source is,

I = V/ (R

_{se }+ R_{L }) , R_{se }and R_{L }are in series ..............(1) If it is to be replaced by a current source then load current must be V

**/**(R_{se }+R_{L }). The total current is T.

Both the resistances will take current proportional to their values.

From the current division in parallel circuit we can write,

............(2)

Now this I

Now this I

_{L }and V/(R_{sh }+R_{L })must be same, so equating (1) and (2), Let internal resistance be, R

_{se }= R_{sh }= R say. Then, V= I x R

_{sh }= I x R or I= V/R

_{sh }**.**I = V/R = V/R

^{.}._{se }

**Key point**: If voltage source is converted to current source, then current source I = V/ R

_{se }with parallel internal resistance equal to R

_{se }.

**Key point**: If current source is converted to voltage source, the voltage source V = I R

_{sh }with sereis internal resistance equal to R

_{sh }.

The direction of current of equivalent circuit source is always from -ve to +ve, internal to the source. While converting current source to voltage source, polarities of voltage is always as +ve terminal at top of arrow and -ve terminal at bottom of arrow, as direction of current is from -ve to +ve, internal to the source. This ensure that current flows from positive to negative terminal in the external circuit.

**Example 1**: Transform a voltage source of 20 volts with an internal resistance of 5 Ω to a current source.

The current of current source is, I = V/Rse = 20/5 = 4 A with internal parallel resistance same as Rse .

Equivalent current source is as shown in the Fig. 3(b).

**Example 2**: Convert the given current source of 50 A with internal resistance of 10 Ω to the equivalent voltage source.

Fig. 4 |

**Solution**: The given values are, I= 50 A and R

_{sh }= 10 Ω.

For the equivalent voltage source,

V= I x = 50 x 10 = 500 V

R

_{se }= R_{sh }= 10 Ω in series. Note the polarities of voltage source, which are such that +ve at top of arrow and -ve at bottom.

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