Question

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

Answer 1

Answer:

The power dissipated by the meter is 1188W

Explanation:

Here we have a circuit constituted with a power source and two resistors in series, we can calculate the power dissipated by the meter using the following formula:

$P=I^2R_m$

We first need to fin the current going through the circuit:

$I=\frac{V}{R}\\where:\\V=voltage\\R=resistance$

$R=R_s+R_m$

because they are connected in series. So:

$I=\frac{12V}{(0.01+0.1)}=109A$

$P=(109)^2*0.1=1188W$