Respuesta :
Current = (voltage) / (resistance)
I(flashlight) = (1.5 v) / (resistance)
----------------------------------------------
I(car battery) = (12 v) / (resistance)
If the resistance is the same in both cases, then
I(flashlight) 1.5
----------------- = ------
I(car battery) 12
The flashlight current is 1/8 (12.5%) as much as the car battery current.
The car battery current is 8 times as much as the flashlight current.
The current is the ratio of voltage and resistance across the circuit. The current flows through the car battery is 8 times as much as the flashlight current.
What is the current?
The current is the rate of flow of electric charge within a circuit. The voltage across an electrical component is needed to make a current flow through it.
Given that the voltage across the flashlight cell is 1.5 V and voltage across a car battery is 12 V. Also the resistance is the same for flashlight cell and car battery.
Let's consider that Vf is the voltage across the flashlight cell and the current flowing through it is I_f. Assume Vc is the voltage across the car battery and Ic is the current flowing through it. The resistance will be the same which is assumed as R.
The current across the flashlight cell is,
[tex]I_f = \dfrac {V_f}{R}[/tex]
[tex]I_f = \dfrac {1.5}{R}[/tex]
The current across the car battery is,
[tex]I_c = \dfrac {V_c}{R}[/tex]
[tex]I_c= \dfrac {12}{R}[/tex]
The ratio of both currents is given as below.
[tex]\dfrac {I_f}{I_c} = \dfrac {\dfrac {1.5}{R}}{\dfrac {12}{R}}[/tex]
[tex]\dfrac {I_f}{I_c} = \dfrac {1.5}{12}[/tex]
[tex]\dfrac {I_f}{I_c} = \dfrac {1}{8}[/tex]
[tex]I_c = 8I_f[/tex]
Hence we can conclude that the current flows through the car battery is 8 times as much as the flashlight current.
To know more about the current, follow the link given below.
https://brainly.com/question/2285102.