How much work a battery does connected to a resistor

AI Thread Summary
A battery with an EMF of 12 volts and internal resistance of 3 ohms is connected to a 21-ohm resistor, allowing a current of 0.5 amps to flow. The potential difference across the resistor is 10.5 volts, leading to a terminal voltage of 10.5 volts. The initial calculation for work done by the battery was 315 joules, but this only accounts for the external resistance. The correct approach involves considering both the external work and the heat generated by the internal resistance, which affects the total work output. Therefore, the total work done by the battery in one minute should be calculated by factoring in both resistances.
Marina1234567
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Homework Statement



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?

Homework Equations



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

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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1 amp x 1 volt = 1 coul/sec x 1 joule/coul = 1 joule/sec = 1 watt
 
Marina1234567 said:
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?
 
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 : W=I^2 R t 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!
 
karkas said:
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 : W=I^2 R t 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
 
Marina1234567 said:

Homework Statement



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?

Homework Equations



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

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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