RESISTANCE OF BALLISTIC GALVANOMETER BY HALF-

DEFLECTION METHOD AND FIND CURRENT SENSITIVITY

PHYSICS PRESENTATION

Compiled By:

1.Abhinav Jain 201/CO/10

2.Abhinav Sangal202/C0/10

3.Abhishek203/CO/10

4.Abhishek Garg204/CO/10

A Ballistic Galvanometer:

Like many other galvanometers, a ballistic galvanometer has a coil which rotates between magnets.

The ballistic galvanometer has the special feature that its rotating coil has a large moment of inertia.

It is used to measure quantity of charge rather than currents, for the large moment of inertia permits the passage of a quantity of charge before the coil moves significantly.

The passage of the charge produces an impulse, a momentary torque , which causes the coil then to swing to some maximum position . Such a galvanometer was often used to standardize capacitors.

The ballistic galvanometer is designed to deflect its indicating needle(or mirror) in a way that is proportional to the total charge passing through its moving coil or to a voltage pulse of short duration.

Any conventional galvanometer may also be employed as a ballistic type, but latter has smaller torque and higher inertia in the coil.

A lamp and scale arrangement:

A 2 Volt Battery

In form of regulated power Supply.

A One Way Key

A Tapping Key

A Ordinary Resistance Box

A Decimal Resistance BoxRange of device:0.1 to 3 Ω

Range of Device: 1 to 2000 Ω

Objective:

To determine resistance of a ballistic galvanometer by half-deflection method and to find out the current sensitivity.

Apparatus Required:

A mega ohm Resistance box.

Range of device:0.1 to 1.0 MΩ

Formulae used for Current sensitivity:

Cg.фo=E.RL/RH.G

Cg.фo=E.RL/RH.R

Circuit Diagram:

Procedure:

Adjust the lamp and scale arrangement to have a clear image of cross wire on the scale.

Make the connections as shown in the diagram but while doing so remove the battery key. The battery key should not be inserted till you have brought in the battery in high resistance by the use of Mega ohm box(otherwise the battery will be connected directly to the galvanometer and it shall get damaged completely.

Put about 0.4 to 0.5 mega ohm resistance in the R.

Note down the deflection when resistance is zero.

Adjust the resistance in R.B. so that the deflection in galvanometer becomes just one half. Note down this value of resistance. This is the resistance of galvanometer.

Repeat steps 4 and 5 by changing RL or RH

In order to calculate the sensitivity record the deflection on the galvanometer for different values of R.

Note down the battery voltage.

Observations:

Distance of scale from galvanometer=1m

Resistance of galvanometer by half deflection and current sensitivity E=4V

S.NO .

RH (Ohms) RL (Ohms)

R (Ohms)

Deflection(ф) (cm)

1. 0.1 X 10^6 1 114 16.5

2. 0.2 x 10^6 2 115 15

3. 0.3 X 10^6 3 115 13

Inference And Result:1. Observed Resistance of the Ballistic

Galvanometer=114.7Ω

2. Current Sensitivity of the ballistic Galvanometer = 1.78x10^-8 A/cm

Percentage Error:

The Percentage error in the observed resistance of the Ballistic galvanometer is 0.26%

Precautions:

1. Do not disturb the setting of the galvanometer. If the setting Is not proper take the help of teacher in leveling the galvanometer

2. See that the light falls normally on the mirror.

3. Do not move the galvanometer without shunting and resting the coil.

4. To damp unnecessary oscillations, use a tapping key.

Thus, current sensitivity can be calculated , provided the resistance of galvanometer is known.

To determine resistance of galvanometer by Half-Deflection method, make the resistance in resistance box R to zero, so that ф=фo Cg.фo=E.RL/RH.G ..…….[4]

Now adjust the resistance in the resistance box R, such that, the deflection in the galvanometer becomes just one half .i.e. it becomes фo . Cg.(фo)=E.RL/RH.(R+G) ………..[5]

From equations (4) and (5), 2G=R+G R=GThis gives, Cg.фo=E.RL/RH.G or Cg.фo=E.RL/RH.R

Theoretical Analysis:

Let,RH – Mega Ohm Resistance Box RL – Decimal Resistance BoxR – Ordinary Resistance BoxE – A battery of known e.m.f.G – Ballistic Galvanometer

If RL is very low resistance as compared to the resistance of galvanometer etc. shunting it by G or R will produce no appreciable change in the current through RL or RH then we have potential difference across RL=E.RL/(RH+RL)Thus, Potential Difference = E.RL/RH as RL<<R ……..[1] And Current in galvanometer = E.RL/RH(R+G) ………[2] Due to this current, suppose the deflection in the galvanometer is ф, then if the current sensitivity of galvanometer is Cg we have, E.RL/RH(R+G)=Cg.ф …….[3]