Batteries in Parallel and series

[Talz1994]

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

 

[gneill]

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.

 

[Talz1994]

oh ok thanks, thread can be closed now.