3) A practical voltage source whose short circuit current is 4 A and open circuit voltage-is 20 V. What is
the voltage across and the value of power dissipated in the load resistance when this source is
delivering a current of 1 A?
Answers
Answer:
In theory the short circuit current for any given voltage is infinite as described by ohms law. Any given voltage divided by zero is infinite. Of course in the real world that can't happen. Your 10 volt source, be it a battery or dynamo etc, has an inherent resistance/impedance. Following ohms law again the voltage in theory collapses to zero when you apply a short circuit to the source. Again in the real world this doesn't happen. So your circuit has a source resistance which limits the current flow to a maximum of 4 amps.
Calculating that (as a purely resistive load) we can calculate the internal resistance which is 10/4= 2.5 add that to your 2 ohm resistor and you have a total resistance of 4.5 ohms. So the current should be 2.222 amps.
It won't be but that should be good enough for your homework. Remember it's Ohm's law. V=IR, I=V/R, R=V/I
Step-by-step explanation:
In theory the short circuit current for any given voltage is infinite as described by ohms law. Any given voltage divided by zero is infinite. Of course in the real world that can't happen. Your 10 volt source, be it a battery or dynamo etc, has an inherent resistance/impedance. Following ohms law again the voltage in theory collapses to zero when you apply a short circuit to the source. Again in the real world this doesn't happen. So your circuit has a source resistance which limits the current flow to a maximum of 4 amps.
Calculating that (as a purely resistive load) we can calculate the internal resistance which is 10/4= 2.5 add that to your 2 ohm resistor and you have a total resistance of 4.5 ohms. So the current should be 2.222 amps.
It won't be but that should be good enough for your homework. Remember it's Ohm's law. V=IR, I=V/R, R=V/I