In a galvanometer setup intended to measure voltages, you often encounter a configuration known as a voltmeter, where a resistor is added in series with the galvanometer to increase its range of measurement.
The basic principle is that the total resistance of the voltmeter (comprising the galvanometer's resistance and the additional series resistor) allows it to handle a higher voltage by limiting the current that flows through the galvanometer. The maximum voltage (V) that can be measured by the galvanometer is determined by Ohm's Law: V = I * R,
Where:
- V is the maximum voltage (40V in this case),
- I is the full scale current rating of the galvanometer,
- R is the total resistance needed so that the galvanometer is not damaged.
Assuming the galvanometer has a known internal resistance (G) and a known full-scale current (I_fullscale), the resistance R required in series can be calculated via the formula:
R = (V / I_fullscale) - G
For this solution, you need either the values of G and I_fullscale or their product (G * I_fullscale). Without those exact specifications provided, it would be imprudent to give an exact numeric answer.
However, if this is a typical example and you have a typical galvanometer with a full-scale current of 50 μA and an internal resistance of 500 Ω, you can compute:
R = (40 / 50 x 10^-6) - 500 = 2000 - 500 = 1500 Ω
Therefore, you would need an additional R = 1990 Ω - 1500 Ω = 490 Ω, meaning the closest possible practical value from your choices is 1990 Ω (including the internal resistance).
If the specific parameters of the galvanometer differ, adjust the calculation accordingly, but the general process is as laid out here.
