1.

A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be :

Answer»

`1A`
0.5A
0.25A
2A

Solution :Equivalent RESISTANCE,

`R.=R+(R_(A)XXS)/(R_(A)+S)`
= `40.8+(480xx20)/(480+20)`
= `40.8+19.2=60.0Omega`
`THEREFORE` Current in the circuit,
`I=V/(R.)=30/60=0.5A`
`therefore` MEASUREMENT of ammeter is 0.5A.


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