Batteries in Parallel and series

AI Thread Summary
The discussion revolves around calculating the power dissipated by a resistor (R2) in a circuit with two batteries connected in series. The user initially calculated the current using the formula I=2E/(2R1+R2) and derived a power value of approximately 0.0714. However, it was pointed out that the calculation mistakenly included all resistances instead of just R2, leading to an incorrect answer. The correct power dissipated by R2, to two significant figures, is 0.064. The thread concludes with the user acknowledging the mistake and indicating that the discussion can be closed.
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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