How would you convert a voltmeter that can measure up to 150 V to an ammeter which can measure current up to 8A ? Resistance of the voltmeter is `300Omega.`

Explanation

Resistance of the voltmeter `R=3000Omega`, maximum voltage `V=150V`.
`:.` Maximum current throgh the voltmeter,
`I_(V)=(V)/(R)=(150)/(300)=0.5A`
Connecting a shunt S in parallel with the volmeter [Fig 1.68] and using this combination in between the points A and B, we shall get an ammeter to measure a maximum current I.
Here, `I=8A`
Now, `I_(S)*S=I_(V)*R`
or, `S=(I_(V))/(I_(S))*R=(I_(V))/(I-I_(V))*R=(0.5)/(8-0.5)xx300=(0.5)/(7.5)xx300`
`=200 Omega`
This is the required value of the shunt-resistance.