Home
Class 12
PHYSICS
Two moving coil meters, M(1) and M(2) ha...

Two moving coil meters, `M_(1) and M_(2)` have the following particulars :
`R_(1)=10 Omega, n_(1)=30`,
`A_(1)=3.6xx10^(-3)m^(2), B_(1)=0.25 T`
`R_(2)=14 Omega, n_(2)=42`
`A_(2)=1.8xx10^(-3)m^(2), B_(2)=0.50 T`
(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of `M_(2) and M_(1)`.

Text Solution

Verified by Experts

Given, `R_(1)=10 Omega, N_(1)=30, A_(1)=3.6xx10^(-3)m^(2), B_(1)=0.25 T`
`R_(2)=14 Omega, n_(2)=42, A_(2)=1.8xx10^(-3)m^(-2), B_(2)=0.50 T, K_(1)=K_(2)`
a) `I=(NAB)/(K)`
`(I_(s2))/(I_(s1))=(N_(2)B_(2)A_(2)K_(1))/(N_(1)B_(1)A_(1)K_(2))=(42xx0.50xx1.8xx10^(-3))/(30xx0.25xx3.6xx10^(-3))=1.4`
b) `V=(NAB)/(KR)`
`(V_(s2))/(V_(s1))=(N_(2)B_(2)A_(2). K_(1)R_(1))/(N_(1)B_(1)A_(1).K_(2)R_(2))=(42xx0.50xx1.8xx10^(-3)xxKxx10)/(Kxx14xx30xx0.25xx3.6xx10^(-3))=1`
Promotional Banner

Topper's Solved these Questions

  • MOVING CHARGES AND MAGNETISHM

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise ADDITIONAL EXERCISES|26 Videos

  • MOVING CHARGES AND MAGNETISHM

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise DAM SURE LAQ|2 Videos

  • MAGNETISM AND MATTER

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise Additional Exercises|19 Videos

  • NUCLEI

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise DAM SURE (VSAQ)|1 Videos

Similar Questions

Explore conceptually related problems

Two moving coil galvanometer, X and Y have coils with resistance 10 Omega and 14 Omega cross-sertional areas 4.8xx10^(-3)m^(2) and 2.4xx10^(-3)m^(2) , number of turns 30 and 45 respectively. They are placed in magnetic field of 0.25 T and 0.50 T respectively. Then, the ratio of their voltage sensitivities and the ratiof of their voltage senstivities are respectively

Two springs having spring constants 25N/m and 16N/m are stretched by forces F_(1) and F_(2) have equal potential energy. The ratio of the forces F_(1) "and" F_(2) is

The following concentrations were obtained for the formation of NH_(3) from N_(2) and H_(2) at equilibrium at 500K . [N_(2)]=1.5xx10^(-2)M . [H_(2)]=3.0xx10^(-2) M and [NH_(3)]=1.2xx10^(-2)M . Calculate equilibrium constant.

Two situations are shown in fig. (a) and (b) In each case m_(1) = 3 kg and m_(2) = 4 kg . If a_(1), a_(2) are the respective accelerations of the blocks in these situations, then the values of a_(1) and a_(2) are respectively [g = 10 ms^(-2)]

For the formation of NH_(3) from N_(2) and H_(2) at 500 K, the concentration of N_(2), H_(2) and NH_(3) at equilibrium are 1. 5xx10^(-2) M and 1.2xx10^(-2)M, respectively. The equilibrium constant for the reverse reaction is

A(a, b) and B(0, 0) are two fixed points. M_(1) is the mid point of AB. M_(2) is the midpoint of bar(AM_(1)), M_(3) is the midpoint of bar(AM_(2)) and so on. Then M_(5) is