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Chapter 25: Problem 23

When a \(9-\mathrm{V}\) battery is temporarily short-circuited, a \(200-\mathrm{mA}\) current flows. What's the battery's internal resistance?

Short Answer

Expert verified
The battery's internal resistance is \(45\) ohms.

Step by step solution

01

Identify the given values

We're given that the battery's voltage (\(V\)) is \(9\) Volts and the current (\(I\)) that flows when it's short-circuited is \(200\) milliamperes or \(0.2\) Amperes.
02

Apply the Ohm's Law

We can solve for the resistance (\(R\)) by rearranging Ohm's law (\(V = IR\)) to solve for \(R\): \(R = V/I\). Substituting the given values, we get \(R = 9 / 0.2\).
03

Find the Resistance

Calculate the value of \(R\) which will give us the internal resistance of the battery. \(R = 9 / 0.2 = 45\) ohms.

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Most popular questions from this chapter

Show that a capacitor is charged to approximately \(99 \%\) of the applied voltage in five time constants \((5 R C)\).

A resistor draws \(1.00 \mathrm{A}\) from an ideal \(12.0-\mathrm{V}\) battery. (a) If an ammeter with \(0.10-\Omega\) resistance is inserted in the circuit, what will it read? (b) If this current is used to calculate the resistance, by what percent will the result be in error?

Show that only half the total energy drawn from a battery in charging an \(R C\) circuit ends up stored in the capacitor. (Hint:What happens to the rest? You'll need to integrate.)

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

Work all circuit problems to two significant figures even if component values are given with one significant figure.What's the emf of a battery that delivers 27 J of energy as it moves \(3.0 \mathrm{C}\) between its terminals?
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