A theoretical Question. I have a battery and a resistor attached in series. Battery is 1V(maximum power it can diliver is 1W) and the resistor is 0.1Ohm(Capable of dissipating 100W). What will be the voltage and current across the resistor.
If we calculate by the power supplied by the battery than V=IR (Ohm's law fails) and otherway around KVL fails. Any help will be appreciated.
Delivered power = 1W at Rsource=Rload
and voltage across Rload = 1/2 Volt
Pload = Vload^2/Rload ---> Rload = Vload^2/Pload ---> Rload = 0.5V^2/1W = 0.25 Ohm
and because in this case Rsource = Rload
---> internal resistance of the battery Rsource = 0.25 Ohm
Delivered power = 1W at Rsource=Rload
and voltage across Rload = 1/2 Volt
Pload = Vload^2/Rload ---> Rload = Vload^2/Pload ---> Rload = 0.5V^2/1W = 0.25 Ohm
and because in this case Rsource = Rload
---> internal resistance of the battery Rsource = 0.25 Ohm
How is the Voltage across Rload =0.5V it will be only in the case when Rload = Rsource but you have calculated Rsource to be 0.25Ohm while Rload is 0.1 Ohm. What will be the current flowing in the circuit.
According to my knowledge I think the Current will be 10A which means that 10W will be required which the battery is unable to supply that is why the voltage drops to 0.1V to keep the power at 1W.
I doubt that it's of much use to guess about the battery. There should be a clarification first.
- Trivial point. The battery voltage is said to be 1 V in post #1 and 12 V in post #5. What's right?
- does max. output 1W define an internal resistance? Although it's a usual specification for RF, it's rarely used for DC sources (except for specification of intrinsic safe circuits).
In this case, you should draw an equivalent circuit that shows the internal battery impedance. It will hopefully help to clarify the circuit behaviour.