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Chapter 25: Problem 21

A defective starter motor draws 300 A from a car's 12 -V battery, dropping the battery terminal voltage to 6 V. A good starter should draw only 100 A. What will the battery terminal voltage be with a good starter?

Short Answer

Expert verified
The battery terminal voltage with a good starter will be 10 V.

Step by step solution

01

Identify Known Variables

From the problem, we know that a faulty motor draws 300 A and drops the voltage to 6 V. A good starter should only draw 100 A. Therefore, the current I1=300 A, voltage V1=6 V, current I2=100 A. We're asked to find the voltage V2.
02

Apply Ohm's Law to Find Resistance

Ohm’s Law is given by \(V = I \times R\) where V is the voltage, I is the current, and R is the resistance. We can rearrange this to find the resistance of the faulty motor, R = V1 / I1 = 6 / 300 = 0.02 \(\Omega\).
03

Calculate Voltage with Good Starter

Now we can apply Ohm’s Law again to find the terminal voltage with a good motor. We now have R and good starter current: I2=100 A. So, V2 = I2 \times R = 100 \times 0.02 = 2 V. But the battery was 12V so battery voltage with good starter is 12-2 = 10 V.

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Most popular questions from this chapter

The voltage across a charging capacitor in an \(R C\) circuit rises to \(1-1 / e\) of the battery voltage in \(5.0 \mathrm{ms}\). (a) How long will it take to reach \(1-1 / e^{3}\) of the battery voltage? (b) If the capacitor is charging through a \(22-\mathrm{k} \Omega\) resistor, what's the capacitance?

What resistance should you place in parallel with a \(56-\mathrm{k} \Omega\) resistor to make an equivalent resistance of \(45 \mathrm{k} \Omega ?\)

You have a battery whose voltage and internal resistance are unknown. Using an ideal voltmeter and an ideal ammeter, how would you determine each of these characteristics?

Show that the quantity \(R C\) has the units of time (seconds).

You have a \(1.0-\Omega,\) a \(2.0-\Omega,\) and a \(3.0-\Omega\) resistor. What equivalent resistances can you form using all three?
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