How Do You Calculate Current in a Half-Full Gasoline Gauge?

[potato987]

could anyone help me with this problem?

Homework Statement

An automobile gasoline gauge is shown schematically in Figure 27-61. The indicator (on the dashboard) has a resistance of 10 . The tank unit is a float connected to a variable resistor whose resistance varies linearly with the volume of gasoline. The resistance is 200 when the tank is empty and 30 when the tank is full.


*Find the current when the tank is half-full*

Homework Equations

V=IR

The Attempt at a Solution

the current when the tank is empty is 12=I*200. so i multiplied that by two to get the answer but it was incorrect. i also tried to plug 15 into the equation since it's half of thirty but that too was wrong

 

[mikelepore]

First let's make a little chart to describe the variable resistor:

200 ohms ... tank empty
? ohms ... tank half-full
30 ohms ... tank full

What number should I replace those question marks with?

 

[Moetasim]

Dear student,

I can assist you with this problem. It seems like you are on the right track by using the equation V=IR. However, there are a few things to consider when solving this circuit indicator problem.

First, we need to determine the total resistance in the circuit. This includes the 10 ohm resistance of the indicator and the variable resistance of the tank unit. Since the tank is half-full, we can assume that the resistance is halfway between 200 ohms and 30 ohms, which would be 115 ohms. Therefore, the total resistance is 125 ohms (10 ohms + 115 ohms).

Next, we can use Ohm's law (V=IR) to solve for the current. We know that the voltage supplied by the battery is 12V, so we can plug that into the equation:

12V = I * 125 ohms

Solving for I, we get I= 12V/125 ohms = 0.096 A or 96 mA. This is the current when the tank is half-full.

I hope this helps you understand the problem better. Let me know if you have any further questions. Good luck with your studies!

Best,

Scientist