How much work a battery does connected to a resistor

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SUMMARY

The discussion centers on calculating the work done by a battery with an EMF of 12.0 volts and an internal resistance of 3 ohms when connected to a 21.0 ohm resistor. The current flowing through the circuit is 0.5 Amps, resulting in a terminal voltage of 10.5 volts across the resistor. The initial calculation of 315 Joules for one minute of operation is incorrect; the correct approach involves using the formula W=I^2 R t, which accounts for the internal resistance, yielding a total work output of 45 Joules over one minute.

PREREQUISITES
  • Understanding of Ohm's Law and circuit analysis
  • Familiarity with the concepts of EMF and terminal voltage
  • Knowledge of electrical work calculations using Joules
  • Basic principles of current flow and resistance in circuits
NEXT STEPS
  • Study the relationship between EMF, terminal voltage, and internal resistance in batteries
  • Learn how to apply the formula W=I^2 R t for calculating work done in electrical circuits
  • Explore the effects of internal resistance on battery performance and efficiency
  • Investigate the thermal effects of current flow through resistive components in circuits
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Students studying electrical engineering, physics enthusiasts, and anyone involved in circuit design or analysis who seeks to deepen their understanding of battery performance and work calculations.

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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