In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 kΩ are used. The figure of merit of the galvanometer is 60 μA/division. In the absence of shunt resistance, the galvanometer produces a deflection of θ = 9 divisions, when current flows in the circuit. The value of the shunt resistance that can cause the deflection of θ/2 is closest to:a)550Ωb)220 Ωc)55Ωd)110ΩCorrect answer is option 'D'. Can you explain this answer?

Question

In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 k&Omega; are used. The figure of merit of the galvanometer is 60 &mu;A/division. In the absence of shunt resistance, the galvanometer produces a deflection of &theta; = 9 divisions, when current flows in the circuit. The value of the shunt resistance that can cause the deflection of &theta;/2 is closest to:a)550&Omega;b)220 &Omega;c)55&Omega;d)110&Omega;Correct answer is option 'D'. Can you explain this answer?

Answer

Ω are connected in series. The galvanometer has a resistance of 50 Ω. A variable resistance R is connected in parallel with the galvanometer. The current through the galvanometer is measured to be half of its full-scale deflection current when the variable resistance is adjusted to 1 kΩ.

To find the resistance of the galvanometer, we can use the principle of the half deflection method. According to this method, when the current through the galvanometer is half of its full-scale deflection current, the resistance in parallel with the galvanometer is equal to the internal resistance of the galvanometer.

Let's denote the resistance of the galvanometer as R_g. We know that the total resistance in the circuit is given by:

1/R_total = 1/R_g + 1/R

Since the current through the galvanometer is half of its full-scale deflection current, we can write:

I = V / (R_g + R) = (1/2) * I_max

where I_max is the full-scale deflection current.

Substituting the given values, we have:

6 / (R_g + R) = (1/2) * I_max

Since the variable resistance R is adjusted to 1 kΩ when the galvanometer shows half deflection, we can write:

R = 1 kΩ = 1000 Ω

Substituting this value, we have:

6 / (R_g + 1000) = (1/2) * I_max

We can rearrange this equation to solve for R_g:

R_g + 1000 = 12 / I_max

R_g = 12 / I_max - 1000

Substituting the given value of I_max, we have:

R_g = 12 / (2 * I_max) - 1000

R_g = 6 / I_max - 1000

Now we can calculate the value of R_g using the given information.

Answer

For deflection of θ/2, current also reduces to I/2 with shunt resistance S.
Hence,

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