To convert a Weston type galvanometer into an ammeter of range 0 1.0 amp. Introduction: A galvanometer can be converted into an ammeter by connecting a very small resistance, called shunt, across its terminals. This resistance can be calculated if the internal resistance of the galvanometer and maximum current which it can draw for full scale deflection are known. Suppose internal resistance of the galvanometer is µRg¶ and current required for full scale deflection is µIg¶ which is very small. If the current passing through the galvanometer exceeds this value, its needle goes out of scale. Now if it is required to measure current µI¶ which is very very large as compared to µIg' then another resistance is connected across the terminals of the galvanometer in such a way that (I ± Ig) current passes through this resistance and only gI current passes through the coil of the galvanometer. So that the resistance shares the extra burden put upon the galvanometer. The extra burden put on the galvanometer is (I ± Ig) amperes. So the value of the resistance connected across the galvanometer should be so calculated that it allows exactly (I ± Ig) amperes of current to pass through it. Hence the galvanometer in combination with calculated resistance can be used to read µI¶ amperes current. Equipment Tools and Supplies:
Weston type galvanometer, high resistance box, low resistance box, two plug keys, ammeter of range 0 1 amp., rheostat, battery, connecting wires and screw gaug e. Lab Procedure: First Step: To find the internal resistance µR g¶ of the galvanometer.
The resistance of the galvanometer can be found by Half deflection method described below: 1. Complete the circuit as shown in the Fig. above µG¶ is a galvanometer, K 1 and K 2 are the plug keys, µR¶ is the high resistance box µS¶ is a low resistance box and B is the battery.
2. Take out 5000 ohms resistance from the resistance box µR¶. Introduce the plug in the key µK 1¶. (Suppose the total number of divisions on the galvanometer scale is 30). Adjust the resistance in µR¶ so that the galvanometer needle resets at its 30th division i.e. it gives full scale deflection. 3. Introduce the plug in key µK 2¶ as well, the deflection in the galvanometer will decrease. Take out a certain resistance µS1¶ from the resistance box µS¶ in such a way that for this value of resistance, the galvanometer gives half scale deflection i.e. the needle of th the galvanometer rests at 15 division of the scale. 4. Take out plug from key K 2 and introduce plug in the K 1. Adjust the resistance R to get a deflection of 20 division in the galvanometer. Close key K 2 also and by adjusting S, th bring the needle back to 10 division. Let the resistance in S be µS2¶. Record and Calculation:
No of Obs.
1 2 3
Deflection in galvanometer
30 divisions 29 28
Second step:
Resistance from the low Resistance box µS¶ for half deflection µS¶ (ohms)
S1 = 140 ohm S2 =140 ohm S3=140 ohm
Internal resistance of galvanometer Rg = S (Ohms)
140 ohm 140 ohm 140 ohm
To determine the current I g for full scale deflection.
1. Measure the voltage of the battery by a voltmeter. Suppo se it is µE¶ volts. 2. Then make the circuit as shown in Fig. below. Take out 5000 ohms resistance from R.B. and introduce the plug in key K. Adjust the resistance, either by increasing it or by decreasing it, so that galvanometer gives full scale deflection. Note the value of resistance µR¶ in R.B. for this deflection and calculate Ig as under:Ig = (E / R + R g) amperes
Third Step: To determine the shunt resistance for conversion of galvanometer into an ammeter of range 0 1.0
This step does not involve any practical set up. It can be calculated theoretically. The Fig. below is used to explain this step only.
Suppose the required value of resistance is µX¶ which, when connected across the galvanometer, gives full scale deflection with the help of battery µB¶ supplying current µI¶ amperes. In this case potential difference across galvanometer is equal to the potential difference across µX¶ i.e. IgRg=(I±Ig)X or X = (IgRg / I ± Ig) Knowing Ig, R g and I, µX¶ can be calculated. (I = I ampere i.e. the range of the ammeter). Fourth Step: It is also theoretical. It involves the determination of length of the wire having resistance µX¶ ohms
We know that the specific resistance µS¶ of the wire is given by S = X (A / L) where µA¶ the area, L the length and X the resistance of the wire to be used Ifµd¶ is the diameter of the wire, then, S = X(d2 / 4L) (A = d2 / 4) 2 or L = (X / S) (d / 4) cm. Knowing x, d and s (from tables) the length of the wire can be calculated. Hint: As the value of µX¶ is very small, the common connecting copper wire should be used for connecting it across the galvanometer. The lengths of the wires of other materials having resistance µX¶ will be very small, even less than the distance between the terminals of the galvanometer
Fifth Step:
Verification of the result:
1. Connect the copper wire of calculated length across the galvanometer. This arrangement, shown in Fig. below converts the galvanometer into an ammeter, whose 30 divisions read one ampere current. So the value of each division of the galvanometer is (1 / 30) ampere.
2. Introduce plug in key µKey¶ adjust the rheostat in such a way that the galvanometer gives full scale deflection. It means that the current read by it is one ampere. If the ammeter A, connected in the circuit, also shows one ampere reading, then the conversion is perfectly all right. 3. Verify it by adjusting the rheostat, so that the galvanometer gives half scale deflection, the reading in the ammeter will also be (1 / 2) ampere.
No of Obs 1
Deflection in Galvanometer ()
High Resistance µR¶ ohms
Maximum Current
30
9260
3.19*10 ^ - 4
2
29
9500
3.11*10 ^- 4
3
28
9800
3.02*10 ^- 4
I = E/R+R g
Ig =3.02*10 ^ - 4 amp
Result: The galvanometer has been converted into an ammeter of range 0 1.0 amp because the difference is zero or very small.
