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प्रश्न
A multirange current meter can be constructed by using a galvanometer circuit as shown in figure. We want a current meter that can measure 10 mA, 100 mA and 1A using a galvanometer of resistance 10 Ω and that prduces maximum deflection for current of 1mA. Find S1, S2 and S3 that have to be used
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उत्तर
A galvanometer can be converted into ammeter by connecting a very low resistance wire (shunt S) connected in parallel with galvanometer. The relationship is given by IgG = (I – Ig) S, where Ig is the range of galvanometer and G is the resistance of galvanometer.
For measuring `I_1 = 10 mA: I_G.G = (I_1 - I_G)(S_1 + S_2 + S_3)`
For measuring `I_2 = 100 mA: I_G(G + S_1) = (I_2 - I_G)(S_2 + S_3)`
For measuring `I_3 = 1 A: I_G(G + S_1 + S_2) = (I_3 - I_G)(S_3)`
Gives `S_1 = 1 Ω, S_2 = 0.1 Ω`
And `S_3 = 0.01 Ω`
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संबंधित प्रश्न
Write the underlying principle of a moving coil galvanometer.
Obtain the expression for current sensitivity of moving coil galvanometer.
Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω, N1 = 30,
A1 = 3.6 × 10–3 m2, B1 = 0.25 T
R2 = 14 Ω, N2 = 42,
A2 = 1.8 × 10–3 m2, B2 = 0.50 T
(The spring constants are identical for the two meters.)
Determine the ratio of
- current sensitivity and
- voltage sensitivity of M2 and M1.
Explain how moving coil galvanometer is converted into a voltmeter. Derive the necessary formula.
Why is it necessary to introduce a radial magnetic field inside the coil of a galvanometer?
With the help of a neat and labelled diagram, explain the principle and working of a moving coil galvanometer ?
Define current sensitivity of a galvanometer.
Why does a galvanometer when connected in series with a capacitor show a momentary deflection, when it is being charged or discharged?
How does this observation lead to modifying the Ampere's circuital law?
Hence write the generalised expression of Ampere's law.
Explain, giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter?
A coil of radius 10 cm and resistance 40 Ω has 1000 turns. It is placed with its plane vertical and its axis parallel to the magnetic meridian. The coil is connected to a galvanometer and is rotated about the vertical diameter through an angle of 180°. Find the charge which flows through the galvanometer if the horizontal component of the earth's magnetic field is BH = 3.0 × 10−5 T.
The AC voltage across a resistance can be measured using a ______.
The current sensitivity of a galvanometer increase by 20%. If its resistance also increases by 25%, the voltage sensitivity will ______.
Assertion (A): On Increasing the current sensitivity of a galvanometer by increasing the number of turns may not necessarily increase its voltage sensitivity.
Reason (R): The resistance of the coil of the galvanometer increases on increasing the number of turns.
Select the most appropriate answer from the options given below:
A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10-divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be ______.
How is current sensitivity increased?
A resistance of 3Ω is connected in parallel to a galvanometer of resistance 297Ω. Find the fraction of current passing through the galvanometer.
To convert a moving coil galvanometer into an ammeter we need to connect a ______.
The figure below shows a circuit containing an ammeter A, a galvanometer G and a plug key K. When the key is closed:
