contestada

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?

Respuesta :

cjmejiab

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

Ohm's law can be expressed as:

[tex]V = I*R[/tex]

Where

V = Voltage

I = Current

R = Resistance

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

[tex]V = 2(1.5) = 3.0V \rightarrow[/tex] Two batteries

[tex]r = (0.255\Omega)+(0.153\Omega)[/tex]

[tex]r = 0.408\Omega[/tex]

[tex]I = 0.6A[/tex]

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

[tex]V = I (R+r)[/tex]

Where R is the resistance of the lamp,

[tex]R = \frac{V}{I}-r[/tex]

[tex]R = \frac{3}{0.6}-0.408[/tex]

[tex]R =4.59\Omega[/tex]

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

[tex]V' = Ir[/tex]

[tex]V' = (0.6)(0.408)[/tex]

[tex]V' = 0.245V[/tex]

Then the fraction of energy trasnferred is

[tex]\frac{V'}{V} = \frac{0.245}{3}*100\\\frac{V'}{V} = 8.16\%[/tex]

Therefore the fraction of the chemical energy transformed is 8.16%