- #1
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