A galvanometer has an internal resistance of 200 Ω and requires a current of 5.0 mA for full-scale deflection. What resistance should be connected in series with this galvanometer to make a voltmeter that reads 25 V at full-scale deflection?

Question

Quizplus · Accepted Answer

The current needed to produce full-scale deflection is given as 5.0 mA. Let's assume this current flows through the entire circuit (i.e. through the internal resistance of 200 Ω and the external resistance we want to find).

According to Ohm's law, the voltage drop across the internal resistance of the galvanometer is:

V_1 = I*R_1
where I is the current and R_1 is the resistance.

Substituting the given values, we get:

V_1 = 5.0 mA * 200 Ω
V_1 = 1.0 V

To read 25 V at full-scale deflection, we need a voltage drop of 25 V across the entire circuit (i.e. both the internal and external resistances). Let R_2 be the external resistance we want to find.

Then, according to Ohm's law again:

V_2 = I*R_total
where V_2 is the voltage drop across R_2 and R_1, and R_total is the total resistance of the circuit.

We want V_2 to be 25 V and I to be 5.0 mA, so:

25 V = 5.0 mA * R_total

R_total = 25 V / 5.0 mA
R_total = 5.0 kΩ

Since we already know that the internal resistance of the galvanometer is 200 Ω, the external resistance we need to add is:

R_2 = R_total - R_1
R_2 = 5.0 kΩ - 200 Ω
R_2 = 4.8 kΩ

Therefore, the best choice is B.

A Galvanometer Has an Internal Resistance of 200 Ω and Requires