An ammeter of resistance `0*80Omega` can measure current upto `1*0A`.
(a) What must be the value of shunt resistance to enable the ammeter to measure current upto `5*0A`?
(b) What is the combined resistance of the ammeter and the shunt?

Explanation

We have, resistance of ammeter, `R_(A) = 0.80` ohm and maximum current across ammeter, `I_(A) = 1.0 A`.
So, voltage across ammeter,
`V=IR = 1.0 xx 0.80 = 0.8 V. `
Let the value of shunt be x .
(i) Resistance of ammeter, with shunt,
`R=(R_(A)x)/(R_(A)+x) = (0.8x)/(0.8+x)`
Current through ammeter, `I=5A`
` therefore ((0.8x)/(0.8+x))xx5=0.8 rArr 0.8x xx 5 = 0.8(0.8 +x)`
`4x=0.64+0.8x`
`x=(0.64)/(3.2)=0.20 Omega`
Thus, the shunt resistance is `0.20 Omega`.
(ii) Combined resistance of the ammeter and the shunt,
`R=(0.8x)/(0.8+x)=(0.8xx0.2)/(0.8+0.2)=(0.16)/(1)=0.16 Omega.`