Question

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.

Short answer

• a) The current in the circuit when the tank is empty is $8.0\times {10}^{-2}A.$
• b)The current in the circuit when the tank is half-full is 0.13A.
• c) The current in the circuit when the tank is full is 0.40A .

Step-by-step solution

The given data

• 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, $\epsilon =12V$.

Understanding the concept of current

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)

a) Calculation of the current when the tank is empty

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.

b) Calculation of the current when the tank is half-full

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.

c) Calculation of the current when the tank is full

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 .