Two voltmeters with a measuring range of 150 V each, but with different internal resistances, are connected

Two voltmeters with a measuring range of 150 V each, but with different internal resistances, are connected in series and connected to a source with a voltage of 220 V. A voltmeter with an internal resistance of 4 kΩ shows a voltage of 104 V. What is the internal resistance of the second voltmeter?

Since we are talking about a series connection, it means that the same current flows through both voltmeters. If 104 volts fell on a voltmeter with a resistance of 4 kOhm, then on the second the drop will be:

220 – 104 = 116 (B).

Current through both according to Ohm’s law:

I = 104/4000 = 52/2000 = 26/1000 = 0.026 (A).

Expressing the required resistance from the same law, we have:

R = 116: 26/1000 = 116 * 1000/26 = 58/13 * 1000 = (52 + 6) / 13 * 1000 = (4 + 6/13) * 1000.

This entry means 4 and a little less than half of kOhm approximately or exactly:

(4 + 6/13) * 1000 = 4461.5385 (Ohm).