Physics, 15.07.2019 17:00 jep88

Agalvanometer with a resistance of 40.0 ω deflects full scale at a current of 2.0 ma. what resistance should be used with this galvanometer in order to construct a voltmeter that can read a maximum of 50 v?

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The resistance needed to be added is R
The Current is 2 ma
The voltage reading is a maximum of 50 volts.
The ma meter has an internal resistance of 40 ohms.

Formula
E = I * R

Givens
E = 50
I = 2 ms
R = R + 40

Solution
E = I * R
I = 2 ma [ 1 amp / 1000 ma] = 0.002 amp
50 = 0.002 * (R + 40) Divide by 0.002
50/0.002 = R + 40
25000 = R + 40 Subtract 40 from both sides.
R = 25000 - 40
R = 24960 Answer

Answer from: isaiahcannon2803

The meter will still deflect full-scale when the current through it is 2 mA.
(That's 0.002 Amp.)
You want that to happen when the meter is connected to two points
with a potential difference of 50v between them.

The total resistance in the meter circuit has to be

R = (voltage) / (current) = (50v) / (0.002 A) = 25,000 Ohms.

You already have 40 ohms in the meter's mechanism.
So technically, if you want to nail it right-on, you'll want
a resistor of exactly 24,960 ohms in series with the galvanometer.

(You can't buy a resistor with this kind of precision. But there are
ways to take a resistor from the store, and increase its resistance
little by little, until you have exactly what you want.)

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Agalvanometer with a resistance of 40.0 ω deflects full scale at a current of 2.0 ma. what resistanc...

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