- #1

Engineer1

Gold Member

- 16

- 1

In this circuit,first there will be voltage drop in the internal resistance of the voltage source.Mostly,the voltage source is assumed ideal in circuits,however,here a practical voltage source is assumed.

So,there will be voltage drop in the internal resistance of the voltage source which will depend on the value of the internal resistance and the remaining voltage will appear at 2 ohms load.The amount of current which flows through that 2 ohms load will depend on the value of the value of the voltage drop across the 2 ohms load.The more the potential difference across the 2 ohms load.the more current will flow.Is this right?

It is not the other way round that the value of the voltage drop in internal resistance is decided by the current which flows through 2 ohms load? First,there should be voltage drop in internal resistance and the remaining voltage should appear at 2 ohms load resistance and the amount of current in 2 ohms will depend on that voltage appearing across 2 ohms.Is that right.

So,there will be voltage drop in the internal resistance of the voltage source which will depend on the value of the internal resistance and the remaining voltage will appear at 2 ohms load.The amount of current which flows through that 2 ohms load will depend on the value of the value of the voltage drop across the 2 ohms load.The more the potential difference across the 2 ohms load.the more current will flow.Is this right?

It is not the other way round that the value of the voltage drop in internal resistance is decided by the current which flows through 2 ohms load? First,there should be voltage drop in internal resistance and the remaining voltage should appear at 2 ohms load resistance and the amount of current in 2 ohms will depend on that voltage appearing across 2 ohms.Is that right.