Contents
With new discoveries and innovations constantly being made, the study of Physics Topics remains a vibrant and exciting field of research.
How is an Ammeter Connected in a Circuit and How is a Voltmeter Connected in a Circuit?
For every electrical instrument the current flowing through it has a maximum permissible limit. If the current exceeds the limit, there is a possibility of damage to the instrument. To protect sensitive instruments like galvanometers and ammeters against possible damages due to heavy current passing through them, an alternative passage is provided as an inbuilt device within these instruments such that a major part of the main current in the circuit passes through this alternative route and a very small part through the instrument. This alternative passage which is nothing but a low resistance connected in parallel with the instrument is called a shunt.
Fig. shows the arrangement of a shunt where a low resistance S (shunt) is connected in parallel with a galvanometer of resistance G.
Let I be the main current, IG be the current through the galva-nometer and IS be the current through the shunt.
So, I = IG + IS …… (1)
Now the equivalent resistance of the combination of the galva-nometer and the shunt = \(\frac{G S}{G+S}\)
From equations (2) and (3) we get, \(\frac{I_G}{I_S}\) = \(\frac{S}{G}\)
Particularly, if S \(\ll\) G, then IG \(\ll\) IS.
The ratio of the main current and the current passing through the galvanometer is known as the multiplying factor or power of a shunt, denoted by n.
∴ \(\frac{I}{I_G}\) = n or, \(\frac{S}{G+S}\) = \(\frac{1}{n}\) [from equation (2)]
or, \(\frac{S}{G}\) = \(\frac{1}{n-1}\) or, S = \(\frac{G}{n-1}\) ….. (4)
Thus the shunt resistance can be adjusted according to the demand of n depending on G.
The shunt has another utility. With the help of the shunt, the galvanometer can almost accurately record the current flowing through the circuit. When a high resistance galvanometer is connected to a circuit, the current drops considerably. But with a shunt which is used in parallel, the equivalent resistance is reduced even below S. So, the main current of the circuit remains practically unchanged. Therefore, it becomes possible to measure the main current almost correctly.
Numerical Examples
Example 1.
To reduce the action of a galvanometer by 25 times, a shunt is added to h. 1f the galvanometer resistance is 1000 Ω, what is the resistance of the shunt?
Solution:
IG = I ᐧ \(\frac{S}{S+G}\) or, \(\frac{I_G}{I}\) = \(\frac{S}{S+G}\)
According to the question, \(\frac{I_G}{I}\) = \(\frac{1}{25}\)
∴ \(\frac{1}{25}\) = \(\frac{S}{S+G}\) or, 25S = S + G or, 24S = 1000
or, S = \(\frac{1000}{24}\) = 41.67 Ω
Example 2.
If a shunt of 1 Ω is connected to a galvanometer of resistance 99 Ω, what fraction of the main current will flow through the galvanometer?
Solution:
The galvanometer resistance G and the shunt resistance S are connected in parallel. So the galvanometer current,
IG = I ᐧ \(\frac{S}{S+G}\) [I = main current]
or, \(\frac{I_G}{I}\) = \(\frac{S}{S+G}\) = \(\frac{1}{1+99}\) = \(\frac{1}{100}\) = 1%
i.e., 1 % of the main current will flow through the galvanometer.
Example 3.
A battery of internal resistance zero is connected to a galvanometer of resIstance 80 Ω and a resistance of 20 Ω in series. A current flows through the galvanometer. If a shunt of 1 Ω resistance is connected to the galvanometer, show that the current that will now flow
through the galvanometer becomes of the previous current.
Solution:
The main current before adding shunt to the circuit
Connection of Ammeter and Voltmeter in a Circuit
An ammeter is used to measure the current and a voltmeter is used to measure the potential difference between any two points of a circuit. Milliainmeter and microammeter are used to measure small currents while milllivoltmeter and microvoltmeter are used to measure small potential differences.
Connection of ammeter: Ammeter is connected in series in an electrical circuit [Fig.], so that it may give the reading of the current when the circuit current passes through it. The main current decreases a little due to the resistance of the ammeter. To overcome this disadvantage a low resistance ammeter should be used.
Connection of voltmeter: A voltmeter of resistance RV is inserted in parallel connection between the two points (here A and B in Fig.) of the circuit across which the potential difference is to be measured. The potential difference across the voltmeter is recorded as that between the two points. But there is a difficulty in this arrangement. Before inserting the voltmeter the resistance between the points A and B was R.
After joining the voltmeter, the resistance between the points A and B is equal to the equivalent resistance of R and RV, which is always less than R. So, the main current increases i.e., the potential difference between A and B also increases. Hence it is desirable that RV should be much greater than R. In that case, the equivalent resistance becomes almost equal to R and the slight increase of the potential difference can be ignored. So, voltmeter is an instrument having a very high resistance and is connected in parallel in an electrical circuit.
Numerical Examples
Example 1.
The Internal resistance of a battery of 100V is 5 Ω. When the emf of the battery is measured by a voltmeter 20% error is found. What Is the resistance of the volt
meter?
Solution:
Suppose, resistance of the voltmeter is R. Its reading is the potential difference of the external circuit.
According to the question, voltmeter reading = IR = 80% of 100V = 80V
∴ Lost volt = Ir = 100 – 80 = 20 V
∴ \(\frac{I R}{I r}\) = \(\frac{80}{20}\) or \(\frac{R}{r}\) = 4
or, R = 4r = 4 × 5 = 20Ω
Example 2.
In a supply line of 100 V there is a resistance of 1000 Ω. In between one terminal of the resistance and its mid point, a voltmeter is connected which gives a reading of 40 V. Determine the resistance of the voltmeter.
Solution:
Suppose, resistance of the voltmeter is R. In Fig., C is the midpoint of the resistance AB.
So, resistance of each portion AC and BC = \(\frac{1000}{2}\) = 500 Ω
Since the reading of the voltmeter =40 V,
∴ VA – VC = 40 V, so VC – VB = 100 – 40 = 60V
Now, main current of circuit,
Example 3.
When a voltmeter of resistance 100 Ω is connected with an electric cell, the reading of the voltmeter is 2 V. When the cell is connected with a resistance of 15 Ω, an ammeter of resistance 1 Ω gives the reading of 0.1 A. Determine the emf of the cell.
Solution:
In the first circuit [Fig.(a)],
From the equations (1) and (2) we have,
2 + 0.02r = 0.1r+ 1.6
or, 0.08r = 0.4
or, r = \(\frac{0.4}{0.08}\) = 5 Ω
so from (1), E = 2 + 0.02 × 5 = 2 + 0.1 = 2.1 V