In a circuit which has a DC power supply, a metal filament lamp and an ammeter connected. Does dividing by 2 the EMF of the DC power supply also halves the current?
1 Answer
No, we expect the current to reduce but by a factor of less than 2.
Explanation:
This question is a little ambiguous, in that we don't know the state of the lamp. Normally, in a circuit where the load is a resistor, cutting the voltage (EMF) in half would also cut the current in half, owing to Ohm's Law:
If the resistance,
However, metal filament in an incandescent lamp changes it's resistance as a function of its temperature - the resistance actually increases as the temperature increases. This is a nice property helps prevent the lamp from from burning out by getting too hot, since the increasing resistance tends to limit the current.
In our question, if the voltage is reduced, this would tend to reduce the current. But this would also cool the filament which would decrease it's resistance. So, we would expect that the current would be reduced, but by a factor of less than 2.
Bonus
The exception here would be if these voltages and currents were so small as to not make the lamp hot enough to be "incandescent" or be glowing. In this case, the change in resistance may be so small as to be insignificant, making the lamp act more like a resistor.
