Two 1.50-v batteries with their positive terminals in thesame direction are inserted in
series into the barrel of a flashlight. one battery has an internalresistance of 0.255 ω , the other
an internal resistance of 0.153 ω .

when the switch isclosed, a current of 600 ma occurs in the lamp.

(a) what is the lamp’s resistance?

(b) what fraction ofthe chemical energy transformed appears as internal energy in the batteries?

To develop this problem it is necessary to collect the concepts related to Ohm's Law.

Ohm's law can be expressed as:

Where

V = Voltage

I = Current

R = Resistance

For our problem the total values of the variables would be given by:

Two batteries

PART A) Whereas the lamp resistance is a surplus greater than the previous resistance given, it must be

Where R is the resistance of the lamp,

PART B) To calculate the percentage difference in the internal potential we start by calculating the internal voltage given by

Then the fraction of energy trasnferred is

Therefore the fraction of the chemical energy transformed is 8.16%

waste

explanation:

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .