Electricity- 2 objects of different power on same battery.


by hasankamal007
Tags: battery, electricity, objects, power
hasankamal007
hasankamal007 is offline
#1
Apr22-12, 01:07 PM
P: 34
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!
Phys.Org News Partner Physics news on Phys.org
Sensitive detection method may help impede illicit nuclear trafficking
CERN: World-record current in a superconductor
Beam on target: CEBAF accelerator achieves 12 GeV commissioning milestone
mfb
mfb is offline
#2
Apr22-12, 03:26 PM
Mentor
P: 10,766
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
hasankamal007 is offline
#3
Apr23-12, 04:46 AM
P: 34
Ok thats 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
BruceW is offline
#4
Apr23-12, 05:43 AM
HW Helper
BruceW's Avatar
P: 3,337

Electricity- 2 objects of different power on same battery.


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 [itex]I[/itex] is the charge going through per second. So current times voltage is the energy lost per second (i.e. the power). This gives us: [itex]P=VI[/itex] We also know for ohmic resistors that [itex]V=IR[/itex] And rearranging these two equations gives:
[tex]P= \frac{V^2}{R} [/tex]
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.


Register to reply

Related Discussions
Battery/general electricity misunderstandings Electrical Engineering 14
why positive or negtive pole of battery can not attract small objects ? Classical Physics 23
Power from a battery PLEASE HELP! Introductory Physics Homework 2
Battery power Classical Physics 21