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Chapter 25: Problem 30

A new mechanic foolishly connects an ammeter with \(0.1-\Omega\) resistance directly across a \(12-\mathrm{V}\) car battery with internal resistance \(0.01 \Omega .\) What's the power dissipation in the meter? (No wonder it gets destroyed!)

Short Answer

Expert verified
The power dissipation in the meter connected to the battery is approximately 1190W.

Step by step solution

01

Determine Total Resistance in the Circuit

In a series circuit, the total resistance \(R_t\) is the sum of all resistances: \(R_t = R_{battery} + R_{ammeter}=0.01 \Omega + 0.1 \Omega = 0.11 \Omega\).
02

Calculate the Current Flowing through the Circuit

Ohm’s Law states that the current \(I\) through a conductor between two points is directly proportional to the voltage \(V\) across the two points, and inversely proportional to the resistance \(R\), expressed as \(I=V/R\).Apply Ohm's Law to find the current: \(I = V/R_t = 12V / 0.11\Omega \approx 109.09A\).
03

Find Power Dissipation in the Ammeter

Power dissipation \(P\) in a component is calculated as \(P=I^2*R\). Substituting \(I = 109.09A\) and \(R = 0.1\Omega\) (resistance of the ammeter), we get: \(P = (109.09A)^2*0.1\Omega \approx 1190W\) which is the power dissipation in the ammeter.

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