Saved Bookmarks
| 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` `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. |
|