Physics

Problem

A mobile phone charger provides a constant output of

5.0\,\text V

. The connecting USB cable has a resistance of

180\,\text{m}\Omega

and must carry a steady current of

2.0\,\text A

when the phone is charging. (i) Calculate the potential difference lost (voltage drop) across the cable. (ii) Determine the voltage actually available at the phone’s terminals during charging. (iii) To ensure that the terminal voltage never falls below

4.9\,\text V

for the same charging current, compute the maximum allowable resistance of the cable. Give your answer in milliohms.

Intuitive explanation

The key insight is that real wires aren't perfect—they have resistance, which causes some of the charger's voltage to be "wasted" heating the cable itself rather than reaching your phone. You'll use Ohm's law ( $V = IR$ ) to find how much voltage drops across the cable's resistance, then subtract that loss from the charger's output to find what actually arrives at the phone. Part (iii) flips the problem: given a minimum acceptable voltage at the phone, work backwards to find the maximum cable resistance that won't cause too much loss.

Intuitive explanation

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