How much work a battery does connected to a resistor


by Marina1234567
Tags: battery, connected, resistor, work
Marina1234567
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#1
Oct21-08, 07:22 PM
P: 1
1. The problem statement, all variables and given/known data

A battery has an EMF of 12.0 volts and internal resistance of 3 ohms
A 21.0 ohm resistance is connected to the battery.
0.5 Amps flow through the battery, hence 0.5 Amps flows through the 21.0 ohm resistor
The potential difference across the 21 ohm resistor is 10.5 Volts, hence the terminal voltage of the battery is 10.5 Volts.

The question is How much work does the battery connected to the 21.0 ohm resisitor perform in one minute?

2. Relevant equations

Voltage = Joule/coloumb
1 amp = coloumb/ second = (6.258 *10^18 electrons) second
1 electron has a charge -1.60*10^-19 coloumbs

3. The attempt at a solution

current = 0.5 Amps = 3.125*10^18 electron/ second = 30 coloumbs pass wire per minute

usisng the definition of EMF (voltage): 30 coloumbs/ minute * 10.5 volts = 315 Joules

however it keeps on telling me that answer is wrong i know its simple i just can't see it right now thanx
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dlgoff
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Oct21-08, 07:43 PM
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1 amp x 1 volt = 1 coul/sec x 1 joule/coul = 1 joule/sec = 1 watt
Satchmo0016
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#3
Mar17-09, 03:25 PM
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Quote Quote by Marina1234567 View Post

usisng the definition of EMF (voltage): 30 coloumbs/ minute * 10.5 volts[/B] = 315 Joules
Using the EMF definition, shouldn't you use EMF and not Terminal voltage?

karkas
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#4
Mar17-09, 03:43 PM
P: 125

How much work a battery does connected to a resistor


My thought on your problem is as follows.
You have a terminal Voltage of 10.5 V, EMF of 12V, Flow of 1/2 A..so pretty much everything you need.

Use the equation for Work on a Resistance : [latex]W=I^2 R t [/latex] where R=r the resistance of the battery. That gives you 45 Joules produced in 60s = 1minute.

I don't take it for granted that I am correct!
Satchmo0016
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Mar17-09, 04:45 PM
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Quote Quote by karkas View Post
My thought on your problem is as follows.
You have a terminal Voltage of 10.5 V, EMF of 12V, Flow of 1/2 A..so pretty much everything you need.

Use the equation for Work on a Resistance : [latex]W=I^2 R t [/latex] where R=r the resistance of the battery. That gives you 45 Joules produced in 60s = 1minute.

I don't take it for granted that I am correct!
Lets try not to mislead him. He had it mostly correct. Since Volt = Joules/Coulomb he already had the current in C/min...

30 C/min * 12 (J/C) = ___J/min
berkeman
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Mar17-09, 05:06 PM
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Quote Quote by Marina1234567 View Post
1. The problem statement, all variables and given/known data

A battery has an EMF of 12.0 volts and internal resistance of 3 ohms
A 21.0 ohm resistance is connected to the battery.
0.5 Amps flow through the battery, hence 0.5 Amps flows through the 21.0 ohm resistor
The potential difference across the 21 ohm resistor is 10.5 Volts, hence the terminal voltage of the battery is 10.5 Volts.

The question is How much work does the battery connected to the 21.0 ohm resisitor perform in one minute?

2. Relevant equations

Voltage = Joule/coloumb
1 amp = coloumb/ second = (6.258 *10^18 electrons) second
1 electron has a charge -1.60*10^-19 coloumbs

3. The attempt at a solution

current = 0.5 Amps = 3.125*10^18 electron/ second = 30 coloumbs pass wire per minute

usisng the definition of EMF (voltage): 30 coloumbs/ minute * 10.5 volts = 315 Joules

however it keeps on telling me that answer is wrong i know its simple i just can't see it right now thanx
Actually I think the only error is due to the question being a little tricky in wording. You calculated the work done by the battery on the external resistance. The battery also heats up due to the current flowing through its internal resistance....


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