Electricity- 2 objects of different power on same battery.

[hasankamal007]

Guess its pretty simple-


Case1-
I connect a bulb of less power to a battery of voltage, say 1.5volts.

Case2-
I connect a bulb of more power to another battery of same voltage, say 1.5volts.

What is difference seen in the two cases? :
• will the battery drainout quicker in case2? But why, if it applies same voltage of 1.5 volts in both cases.
• or does a battery simply applies relatively more voltage in case2?
• or anything else?

Please clarify. Thanks a lot!

 

[mfb]

Bulb 2 will have a lower resistance (if designed for the same voltage) and therefore drain battery 2 quicker.

If you consider the internal resistance of the battery, the voltage at bulb 2 will be a bit lower (as a higher voltage drops occurs in the battery).

 

[hasankamal007]

Ok that's what I had accepted. But an you answer why it'll drain quicker in case 2? even If it holds same voltage 1.5 in both cases?
Maybe because the battery uses more of its chemicals to maintain same voltage?
(mfb thanks for replying).

 

[BruceW]

mfb is right. Maybe it needs a bit more explanation? Voltage is energy per charge (since we are ignoring the effect of magnetic fields). And we can also say that current $I$ is the charge going through per second. So current times voltage is the energy lost per second (i.e. the power). This gives us: $P=VI$ We also know for ohmic resistors that $V=IR$ And rearranging these two equations gives:
$$P= \frac{V^2}{R}$$
So the power used is inversely proportional to the resistance (since voltage is the same in both cases). To put all this into an intuitive explanation: A greater current means more power because more charge carriers are moving through the potential difference. And to get a greater current, we can use a resistor with less resistance.

 

[sardar]

I can provide some clarification on the difference between the two cases. In both cases, the battery is supplying a constant voltage of 1.5 volts to the bulbs. However, the difference lies in the amount of power (measured in watts) that each bulb requires.

In Case 1, the bulb has a lower power requirement, so it will draw less current (measured in amps) from the battery. This means that the battery will take longer to drain out because it is supplying a smaller amount of current to the bulb.

In Case 2, the bulb has a higher power requirement, so it will draw more current from the battery. This means that the battery will drain out quicker because it is supplying a larger amount of current to the bulb.

So, in short, the difference between the two cases is the amount of current being drawn from the battery, not the voltage. The battery is still supplying the same voltage in both cases, but the amount of power required by the bulb determines the rate at which the battery will drain.

I hope this helps to clarify the concept of power and current in relation to batteries and electricity.