Suppose, we are having a complex network of different voltage sources and resistors
Now, say we have to calculate current through any particular branch in that network
If we now concentrate on that branch only, then rest of the network can be considered
as a single voltage source whose terminal voltage is nothing but the voltage that appears
across the branch for the existing complex network. Internal resistance of that voltage source will be the equivalent resistance of the network
that appears between the terminals of the branch. If we go through one example, that will be clear to us
Say, in the network, we have to calculate current through the resistor RL across the
terminals A and B. For that, we first remove the resistor RL from the terminals A and B of the network
Now if we observe the network carefully, the voltage that appears across the terminal AB
is nothing but the voltage drop across the 4 ohms resistor as shown
Here in the network, this 2 ohms resistor and 4 ohms resistor are series connected
Hence equivalent resistance of this combination will be 6 ohms. Now this equivalent 6 ohms is in parallel with another 6 ohms resistor
Hence resultant resistance of this combination will be 6 by 2 that is 3 ohms
This 3 ohms resistance then again is in series with another 3 ohms resistance
Hence resistance offered by the network across the voltage source is 3 plus 3 that is 6 ohms
Hence at this situation, the total current supplied by the source to the network is 6
volts by 6 ohms that is 1 ampere. This 1 ampere is then divided into two equal parts of 0.5 ampere as shown
Hence current flow through the 4 ohms resistor is 0.5 ampere. That means voltage drop across 4 ohms resistor will be 4 into 0.5 that is 2 volts
This 2 volt is actually offered by the network across the terminals A and B
Hence the total network can be considered as a single source of 2 volts
For calculating equivalent resistance of the network across the terminal A and B, we
have to first replace the voltage source in the network by its internal resistance
Here in this example, the only voltage source of 6 volts is taken as ideal
So we will replace it by a short circuit. After replacing the source by short circuit, we will look at the network from points A
and B. Here in the network, this 3 ohms resistor and 6 ohms resistor are parallel
Hence equivalent resistance of this combination will be 3 into 6 divided by 3 plus 6 that
is 2 ohms. Now this equivalent 2 ohms is in series with another 2 ohms resistor
Hence resultant resistance of this combination will be 2 plus 2 that is 4 ohms
This 4 ohms resistance then again is in parallel with another 4 ohms resistance
Hence resistance offered by the network is 4 by 2 that is 2 ohms
Hence open circuit voltage and resistance offered by the network across the points A
and B are 2 volts and 2 ohms respectively. Thus this network can be replaced by a single voltage source of 2 volts with internal resistance
of 2 ohms connected in series with it