Batteries in Parallel and series

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SUMMARY

The discussion focuses on calculating the power dissipated by a resistor (R2) in a circuit with two batteries connected in series, each with an EMF of 10V and an internal resistance of 300 ohms. The user initially calculated the power using the formula P=I^2.R but included the total resistance instead of just R2. The correct power dissipated by R2, after adjusting the calculations, is determined to be 0.064 watts to two significant figures.

PREREQUISITES
  • Understanding of electrical circuits and Ohm's Law
  • Familiarity with the concepts of EMF and internal resistance
  • Knowledge of power calculations in electrical circuits
  • Proficiency in algebra for manipulating equations
NEXT STEPS
  • Review the principles of series and parallel circuits
  • Learn about calculating power in electrical circuits using P=I^2.R and P=V^2/R
  • Study the effects of internal resistance on circuit performance
  • Explore advanced circuit analysis techniques, such as Thevenin's and Norton's theorems
USEFUL FOR

Students studying electrical engineering, educators teaching circuit theory, and anyone interested in understanding power dissipation in electrical circuits.

Talz1994
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Homework Statement


You are given two circuits with two batteries of emf "E" and internal resistance R1 each. Circuit A has the batteries connected in series with a resistor of resistance R2, and circuit B has the batteries connected in parallel to an equivalent resistor.
Note that the symbol E should be entered in your answers as EMF.

What is the power dissipated by the resistor of resistance R2 for circuit A, given that
EMF=10 R1=300 ohms R2=5000 Ohms

Calculate the power to two significant figures.


Homework Equations


P=I^2.R
P=V^2/R


The Attempt at a Solution



well i used the second equation, and from previous question worked out my current in algebraic equation,
I=2E/2R1+R2 so I= 2x10/2x300+5000 =1/280 R=5000+2x300= 5600

P=I^2.R

P=(1/280)^2.5600= 0.07142857143

how ever the correct answer to 2.s.f is 0.064
 
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Talz1994 said:

Homework Statement


You are given two circuits with two batteries of emf "E" and internal resistance R1 each. Circuit A has the batteries connected in series with a resistor of resistance R2, and circuit B has the batteries connected in parallel to an equivalent resistor.
Note that the symbol E should be entered in your answers as EMF.

What is the power dissipated by the resistor of resistance R2 for circuit A, given that
EMF=10 R1=300 ohms R2=5000 Ohms

Calculate the power to two significant figures.


Homework Equations


P=I^2.R
P=V^2/R


The Attempt at a Solution



well i used the second equation, and from previous question worked out my current in algebraic equation,
I=2E/2R1+R2 so I= 2x10/2x300+5000 =1/280 R=5000+2x300= 5600

P=I^2.R

P=(1/280)^2.5600= 0.07142857143

how ever the correct answer to 2.s.f is 0.064

They're only looking for the power dissipated by R2. You've included all the resistance in your R.
 
oh ok thanks, thread can be closed now.
 

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