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प्रश्न
Explain, giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter?
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उत्तर
(i) In converting a galvanometer into a voltmeter, a very high suitable resistance is connected is series to its coil. So, the galvanometer gives full scale deflection.
(ii) In converting a galvanometer into an ammeter, a very small suitable resistance is connected in parallel to its coil. The remaining pair of the current i.e. (I − Ig) flows through the resistance. Here I = Circuit current
Ig = Current through galvanometer
संबंधित प्रश्न
The combined resistance of a galvanometer of resistance 500Ω and its shunt is 21Ω. Calculate the value of shunt.
The combined resistance of a galvanometer of resistance 500Ω and its shunt is 21Ω. Calculate the value of shunt.
- A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
- Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
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How will you convert a moving coil galvanometer into a voltmeter?
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How does this observation lead to modifying the Ampere's circuital law?
Hence write the generalised expression of Ampere's law.
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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 ______.
