Question

The starting motor of a car is turning too slowly, and the mechanic has to decide whether to replace the motor, the cable, or the battery. The car’s manual says that the12Vbattery should have no more than$\mathbf{0}\mathbf{.}\mathbf{020}\mathbf{\Omega }$ internal resistance; the motor should have no more than $\mathbf{0}\mathbf{.}\mathbf{200}\mathit{\Omega }$resistance, and the cable no more than $\mathbf{0}\mathbf{.}\mathbf{040}\mathit{\Omega }$resistance. The mechanic turns on the motor and measures 11.4Vacross the battery, a 3.0Vcross the cable, and a current of 50A. Which part is defective?

Short answer

The cable part of the motor is defective.

Step-by-step solution

The given data

• a) The potential of the battery should have no more internal resistance $r_1=0.020\Omega$, is $\epsilon =12V$.
• b)The motor should have no more resistance,$R_1=0.200\Omega$.
• c) The cable should have no more resistance,$R_2=0.040\Omega$ .
• d)Potential measured across the battery, V_battery =11.4V .
• e) Potential measured across the cable, V_cable= 3.0V.
• f) The current of the motor, i=50A.

Understanding the concept of resistance

In the given problem, it can be seen that there is a limit set to every resistance value of the different parts of the motor. Thus, if the resistance value when exceeds the limit value, that part stops working due to the over resistance barrier to the current flow.

Formula:

The voltage equation using Ohm’s law,

V=IR (i)

HereR is the resistance and I is the current.

Calculation to check the defective part of the motor

All three circuit elements are connected in series, so the current is the same in all of them. The battery is discharging, so the potential drop across the terminals is given using equation (i) as follows:

$V_{battery}=\epsilon -ir$ (a)

Now, the internal resistance of the battery can be given using equation (a), and the given data is as follows:

$r=\frac{\epsilon -V_{battery}}{i}$

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

$$\begin{gathered} r=\frac{12V-11.4V}{50A} \\ =0.012\Omega \end{gathered}$$

The above value is less than that given in the problem, that is$0.012\Omega$. Thus the battery is working.

Now, the resistance value of the motor can be given using the given data in equation (i) as follows:

$R_{motor}=\frac{V_{battery}-V_{cable}}{i}$

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

$$\begin{gathered} R_{motor}=\frac{11.4V-3.0V}{50A} \\ =0.17\Omega \end{gathered}$$

The above value is less than that given in the problem, that is$0.20\Omega$. Thus, the motor is also working.

Now, the resistance value of the cable can be given using the given data in equation (i) as follows:

$$\begin{gathered} R_{cable}=\frac{3.0V}{50A} \\ =0.060\Omega \end{gathered}$$

The above value is greater that given in the problem that is$0.040\Omega$. Thus, the cable is also defective.

Hence, the cable part is defective.