Correct Answer - (a) Shunt resistance `4*0xx10^-3Omega`
(b) series resistance `3988Omega`.
(A) For conversion into ammeter:
`G=12*0Omega`, `I_g=2*5mA=2*5xx10^-3A`,
`I=7*5A`
`S=(I_gG)/(I-I_g)=((2*5xx10^-3)xx12)/(7*5-(2*5xx10^-3))=(2*5xx12xx10^-3)/(7*4975)`
`=4*0xx10^-3Omega` in parallel
(b) For conversion into voltmeter
`G=12*0Omega, I_g=2*5xx10^-3A, V=10V`
`R=V/I_g-G=(10)/(2*5xx10^-3)-12`
`=4000-12=3988Omega` in series