Chapter 28: Q. 65 (page 794)

A 12 V car battery dies not so much because its voltage drops but because chemical reactions increase its internal resistance. A good battery connected with jumper cables can both start the engine and recharge the dead battery. Consider the automotive circuit of FIGURE P28.65.
a. How much current could the good battery alone drive through the starter motor?
b. How much current is the dead battery alone able to drive through the starter motor?
c. With the jumper cables attached, how much current passes through the starter motor?
d. With the jumper cables attached, how much current passes through the dead battery, and in which direction?

Short Answer

Expert verified

a. The current is $200 A$

b. The current is $14.5 A$

c. The current is role="math" localid="1648844396007" $199 A$

d. role="math" localid="1648844482262" $3.93 A$ pass through the dead battery, downwards from its positive towards its negative pole.

Step by step solution

Part (a) Step 1 : Given Information.

We need to find how much current could the good battery alone drive through the starter mode.

Part (a) Step 2 : Simplify

Only good battery is connected to the starter motor.

The total resistance in the circuit is

$r_{g} + R = 0.06 Ω$ .

The current is then

$I_{g} = \frac{12 V}{0.06 Ω} = 200 A .$

Part (b) Step 1 : Given Information.

Need to find dead battery alone able to drive through the starter motor.

Part (b) Step 2 : Clarification

Only the dead battery is connected to the starter motor.

The total resistance in the circuit is

$r_{d} + R = 0.55 Ω$

The current is then

$I_{d} = \frac{8 V}{0.55 Ω} = 14.5 A$

Part (c) Step 1 : Given Information

Need to get current passes through the starter motor.

Part (c) Step 2 : Simplification

Set we have to apply Kirchoff's law to solve for I.

Take the loop abcdand go around it in clockwise direction. According to second Kirchoff's law we have

$ε_{g} - I_{1} r_{g} - I R = 0 . (e q - 1)$

Now, take the loop ecdfand go around it in the clockwise direction. Second Kirchoff's law says

$ε_{d} - I_{2} r_{d} - I R = 0 . (e q - 2)$

First Kirchoff's law for the vertex esays

$I = I_{1} + I_{2 .} (e q - 3)$

Easy way to solve the system of equation for I the following. Multiply (eq-1) with $r_{d}$ and (eq-2) with $r_{g}$

localid="1648851702016" $r_{d} ε_{g} - I_{1} r_{g} r_{d} - I R r_{d} = 0, r_{g} ε_{d} - I_{2} r_{g} r_{d} - I R r_{g} = 0 .$

Now add these two equations together

localid="1648851712246" $r_{d} ε_{g} + r_{g} ε_{d} - (I_{1} + I_{2}) r_{g} r_{d} - I R (r_{d} + r_{g}) = 0 .$

Use (eq-3)to get

localid="1648851721615" $r_{d} ε_{g} + r_{g} ε_{d} - I r_{g} r_{d} - I R (r_{d} + r_{g}) = 0$

i.e.

localid="1648851730214" $r_{d} ε_{g} + r_{g} ε_{d} - I (r_{g} r_{d} + R (r_{g} + r_{d})) = 0$

This is now easily solved for I :

localid="1648851738847" $I = \frac{r_{d} ε_{g} + r_{g} ε_{d}}{r_{g} r_{d} + R (r_{d} + r_{g})}$

Putting the numerical values we find

localid="1648851756624" $I = 199 A$

Part (d) Step 1 : Given Information.

Need to find jumper cables attached current passes through the dead battery in which direction.

Part (d) Step 2 : Explanation

We have supposed that flows upwards (from - to + inside the battery). This supposition was made so that we can write down equation that we have derived from Kirchoff's 1st and 2nd law.

However, $I_{2}$ might turn out to be negative meaning that the physical current actually flows in the opposite direction. To see if this is the case we express $I_{2}$ from (eq-2):

$I_{2} = \frac{ε_{d} - I R}{r_{d}} .$

Substituting numerical values we find

$I_{2} = - 3.93 A$

Therefore, the current as the magnitude of $3.93 A$ and it passes downwards, from + to - inside the battery, opposite of what we have supposed in part c.