Question: An automobile gasoline gauge is shown schematically in Fig. 27-74. The indicator (on the dashboard) has a resistance of$10\Omega$. The tank unit is a float connected to a variable resistor whose resistance varies linearly with the volume of gasoline. The resistance is$140\Omega$when the tank is empty and$20\Omega$when the tank is full. Find the current in the circuit when the tank is (a) empty, (b) half-full, and (c) full. Treat the battery as ideal.

• a)Resistance of the indicator,$R_{Indicator}=10\Omega$ .
• b)Resistance when tank is empty,$R=140\Omega$.
• c)Resistance when tank is full,$R_{full}=20\Omega$ .
• d) Emf of the battery as shown in the figure, $\epsilon =12V$.

The current I flowing through a body is determined by the voltage V in the circuit divided by the resistance ratio, which is given by both the internal and external resistances.

Formula:
The voltage equation using Ohm’s law, V= IR (i)

Now, the current flowing within the tank when the tank is empty is given using the given data in equation (i) as follows:

$I=\frac{\epsilon }{R_{Indicator}+R_{empty}}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} I=\frac{12V}{10\Omega +140\Omega } \\ =8.0\times {10}^{-2}A \end{gathered}$$

Hence, the value of current is$8.0\times {10}^{-2}A$35.

The equivalent resistance of case of half-full or half empty can be given as:

$R_{eq}=\frac{R_{full}+R_{empty}}{2}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} R_{eq}=\frac{20\Omega +140\Omega }{2} \\ =80\Omega \end{gathered}$$

Now, the current flowing within the tank when the tank is empty is given using the given data in equation (i) as follows:

$I=\frac{\epsilon }{R_{Indicator}+R_{eq}}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} l=\frac{12V}{10\Omega +80\Omega } \\ =0.13A \end{gathered}$$

Hence, the value of current is 0.13 A.

Now, the current flowing within the tank when the tank is full is given using the given data in equation (i) as follows:

$l=\frac{\epsilon }{R_{Indicator}+R_{full}}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} l=\frac{12V}{10\Omega +20\Omega } \\ =0.40A \end{gathered}$$

Hence, the value of current is 0.40A .