What is the bulb's resistance?

+2 votes

Two 1.50-V batteries—with their positive terminals in the same direction—are inserted in series into the barrel of a flashlight. One battery has an internal resistance of 0.380 Ω, the other an internal resistance of 0.150 Ω. When the switch is closed, a current of 600 mA occurs in the lamp.

(a) What is the bulb's resistance?
Ω

(b) What fraction of the chemical energy transformed appears as internal energy in the batteries?
%
asked Feb 4, 2012 in Physics by yoshi ~Expert~ (919 points)
recategorized Feb 5, 2012 by pokemonmaster
    

1 Answer

+3 votes
 
Best answer
First recall that resistance is:

R = V / I (*where R=resistance, V=voltage, I=current)

So use this formula to calculate the total resistance of the circuit.

V = 3.00 V (because 1.5+1.5 = 3 V)

I = 600mA = 600*10^-3 A = 0.6 A

And so total resistance is

 

R = 3.00 V / 0.6 A  = 5 Ohms

Now to find the bulbs resistance we just subtract the batteries resistance from the total resistance.
answered Feb 7, 2012 by pokemonmaster ~Expert~ (3,856 points)
selected Feb 9, 2012 by yoshi

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