A galvanometer coil has a resistance of `15Omega` and the meter shows full scale deflection for a current of `4mA`. How will you convert the meter into an ammeter of range 0 to 6A?

A galvanometer coil has a resistance of `15Omega` and the meter shows

1.	A galvanometer coil has a resistance of `15Omega` and the meter shows full scale deflection for a current of `4mA`. How will you convert the meter into an ammeter of range 0 to 6A?
Answer» Here, `G=15Omega`, `I_g=4mA=4xx10^-3A`, `S=?`, `I=6A` From `S=(I_g.G)/(I-I_g)=(4xx10^-3xx15)/(6-4xx10^-3)=(6xx10^-2)/(5.996)=10^-2ohm` This low resistance, called shunt resistance, is to be connected in parallel with the galvanometer.

Discussion

No Comment Found

Related InterviewSolutions

A current carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin, joined together. Find the direction and magnitude to `vecB` at the origin.
A long straight wire carries a current of `2A`. An electron travels with a velocity of `4xx10^4ms^-1` parallel to the wire `0*1mm` from it, and in a direction opposite to the current. What force does the magnetic field of current exert on the moving electron. Charge on electron `=1*6xx10^(-19)C`.
The variation of magnetic susceptibility `chi` with the temperature T of a ferromagnetic material can be plotted asA. B. C. D.
A wire of length L metre carrying a current of I ampere is bent in the form of a circle. Find its magnetic moment.
A wire of length `L metre ` , carrying a current `I `ampere is bent in the form of a circle . The magnitude of its magnetic moment is …………. ` MKS units ` .A. `(IL)/(4pi)`B. `(IL^(2))/(4pi)`C. `(I^(2)L^(2))/(4pi)`D. `(I^(2)L)/(4pi)`
A wire of length `L metre ` , carrying a current `I `ampere is bent in the form of a circle . The magnitude of its magnetic moment is …………. ` MKS units ` .
A long wire carries a steady curent . It is bent into a circle of one turn and the magnetic field at the centre of the coil is `B`. It is then bent into a circular loop of `n` turns. The magnetic field at the centre of the coil will beA. nBB. `n^(2)B`C. 2nBD. `2n^(2)B`
A long solenoid has a radius `a` and number of turns per unit length is `n`. If it carries a current i, then the magnetic field on its axis is directly proportional toA. `ani`B. `ni`C. `(ni)/a`D. `n^(2)i`
Two concentric circular coils of ten turns each are situated in the same plane. Their radii are `20` and `40 cm` and they carry respectively `0.2` and `0.3` ampere current in opposite direction. The magnetic field in `Wb//m^(3)` at the centre isA. `35/4 mu_(0)`B. `(mu_(0))/80`C. `7/80 mu_(0)`D. `5/4 mu_(0)`
A particle of mass `m` and charge `q` moves with a constant velocity `v` along the positive `x` direction. It enters a region containing a uniform magnetic field `B` directed along the negative `z` direction, extending from `x = a` to `x = b`. The minimum value of `v` required so that the particle can just enter the region `x gt b` isA. `qb B//m`B. `q(b-a)B//m`C. `qa B//m`D. `a(b+a)B//2m`

A galvanometer coil has a resistance of `15Omega` and the meter shows full scale deflection for a current of `4mA`. How will you convert the meter into an ammeter of range 0 to 6A?

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment