Question

A galvanometer is used to detect or/and measure small currents in an electrical circuit. It essentially works on the fact that a current-carrying coil experiences a deflecting torque when placed in a magnetic field. This deflection in the coil can be measured and it is related to the current flowing in the coil, the number of turns in the coil, area of the coil and the magnetic field. A hair spring attached to the coil provides a counter torque and helps in measuring the deflection. A galvanometer can be converted to an ammeter or a voltmeter of desired range by using suitable resistances.

(I) The torque on the coil remains constant irrespective of the coil’s orientation during rotation due to ______.

1. use of soft iron core which increases the magnetic field.
2. radial magnetic field.
3. hair spring which provides the counter torque.
4. eddy current in the iron core which causes damping.

(II) The best way to increase current sensitivity of a galvanometer is by ______.

1. increasing number of turns of the coil.
2. increasing area of coil and magnitic field strength.
3. decreasing area of coil and magnetic field strength.
4. increasing torsional constant of the hair spring.

(III) A moving coil galvanometer has a coil with area of cross-section 4.0 × 10⁻³ m² and number of turns 50. The coil is rotating in a magnetic field of 0.25 T. The torque acting on the coil when a current of 5 A passes through it is ______.

1. 1.0 N m
2. 2.0 N m
3. 0.50 N m
4. 0.25 N m

OR

A galvanometer coil has a resistance of 15 Ω and the meter shows full scale deflection for a current of 3 mA. The value of resistance required to convert it into a voltmeter of range (0-12 V) is ______.

1. 4015 Ω
2. 3985 Ω
3. 415 Ω
4. 385 Ω

(IV) A galvanometer with coil of resistance 20 Ω shows full scale deflection for a current of 5 mA. To convert it into an ammeter of range (0-10 A), a resistance of ______.

1. 0.05 Ω should be connected in series with it.
2. 0.05 Ω should be connected in parallel with it.
3. 0.01 Ω should be connected in parallel with it.
4. 0.01 Ω should be connected in series with it.

Answer 1

(I) The torque on the coil remains constant irrespective of the coil’s orientation during rotation due to radial magnetic field.

Explanation:

In a moving coil galvanometer, the deflecting torque acting on the coil is given by:

τ = nBIA sin θ

For accurate measurement, torque must be proportional only to current and not depend on angle.

Problem with a uniform magnetic field. If the magnetic field is uniform, the torque depends on sin θ, so it changes with coil orientation. This makes the scale non-linear.

Use of radial magnetic field. In a radial magnetic field, the magnetic field lines remain perpendicular to the plane of the coil. Thus, the angle between the magnetic field and the coil’s normal stays constant:

θ = 90°

⇒ sin θ = 1

So torque becomes:

τ = NBIA

Which is independent of coil orientation.

(II) The best way to increase current sensitivity of a galvanometer is by increasing area of coil and magnetic field strength.

Explanation:

Current sensitivity:

S = `theta/I`

= `(n B A)/k`

It increases with a larger area, a stronger magnetic field, more turns and a smaller torsional constant.

(III) A moving coil galvanometer has a coil with area of cross-section 4.0 × 10⁻³ m² and number of turns 50. The coil is rotating in a magnetic field of 0.25 T. The torque acting on the coil when a current of 5 A passes through it is 0.25 N m.

Explanation:

Given: Number of turns (n) = 50

Magnetic field (B) = 0.25 T

Current (I) = 5 A

Area (A) = 4 × 10⁻³ m²

Torque on coil (τ) = nBIA

= 50 × 0.25 × 5 × 4 × 10⁻³

= 0.25 N m

OR

A galvanometer coil has a resistance of 15 and the meter shows full scale deflection for a current of 3 mA. The value of resistance required to convert it into a voltmeter of range (0-12 V) is 3985 Ω.

Explanation:

Given: G = 15 Ω

I_g = 3 × 10⁻³ A

V = 12 V

R_s = `V/I_g - G`

= `12/(3 xx 10^-3) - 15`

= 4000 − 15

= 3985 Ω

(IV) A galvanometer with coil of resistance 20 Ω shows full scale deflection for a current of 5 mA. To convert it into an ammeter of range (0-10 A), a resistance of 0.01 Ω should be connected in parallel with it.

Explanation:

Given: G = 20 Ω

I_g = 5 × 10⁻³ A

I = 10 A

For converting a galvanometer into an ammeter, use a shunt resistance:

S = `(I_g G)/(I - I_g)`

= `((5 xx 10^-3) xx 20)/(10 - (5 xx 10^-3))`

= `(0.005 xx 20)/(10 - 0.005)`

= `0.1/9.995`

≈ 0.01 Ω

A shunt is always connected in parallel.