Question

A twenty-five volt battery, an ammeter, and a 5 ohm resistor are connected in series. Assume the resistor obeys Ohm’s law. There could be other circuit elements in the circuit. If the ammeter reads 1.9 A, are there other elements in the circuit? If so, what are they? Explain your reasoning using Kirchoff’s Laws and draw the circuit in the space below.

Answer 1

If there is nothing else in the circuit, then current should be $=V/R=25/5=5A$

This means there are other elements in the circuit. Let their resistance be r.

Now, I=$V/(R+r)=1.9=25/(5+r)$

$r=8.158\Omega$

As voltage across battery is 25v while voltage drop in new circuit is 9.5, so there should be another resistor of resistance 8.158