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

The power dissipated by the meter is 1188W

Step-by-step 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=(V)/(R)\\where:\\V=voltage\\R=resistance`

`R=R_s+R_m`

because they are connected in series. So:

`I=(12V)/((0.01+0.1))=109A`

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