A circuit contains an ammeter, a battery of 45 V and resistance 40.8 ohm, all connected in series. If the ammeter has a coil resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be ampere.

Question

A circuit contains an ammeter, a battery of 45 V and resistance 40.8 ohm, all connected in series. If the ammeter has a coil resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be ampere. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ae89fb16e5039a62148fa87c0cb2eccc-.png" /> Here combination of (480 Ω || 20 Ω ) is in series with 40.8 Ω R text eff =40.8+(480 × 20/480+20) =40.8+19.2=60 Ω I =( V text eff / R text eff )=(45/60)=0.75 A

Q. A circuit contains an ammeter, a battery of $45 V$ and resistance $40.8 o hm$ , all connected in series. If the ammeter has a coil resistance $480 o hm$ and a shunt of $20 o hm$ , the reading in the ammeter will be _______ ampere.

Here combination of $(480 Ω∣∣20 Ω$ ) is in series with $40.8 Ω$
$R_{eff} = 40.8 + \frac{480 \times 20}{480 + 20}$
$= 40.8 + 19.2 = 60 Ω$
$I = \frac{V _{eff}}{R _{eff}} = \frac{45}{60} = 0.75 A$

Q. A circuit contains an ammeter, a battery of 45V and resistance 40.8ohm, all connected in series. If the ammeter has a coil resistance 480ohm and a shunt of 20ohm, the reading in the ammeter will be _______ ampere.

Here combination of (480Ω∣∣20Ω ) is in series with 40.8Ω Reff ​=40.8+480+20480×20​ =40.8+19.2=60Ω I=Reff ​Veff ​​=6045​=0.75A

Q. A circuit contains an ammeter, a battery of $45 V$ and resistance $40.8 o hm$ , all connected in series. If the ammeter has a coil resistance $480 o hm$ and a shunt of $20 o hm$ , the reading in the ammeter will be _______ ampere.

Here combination of $(480 Ω∣∣20 Ω$ ) is in series with $40.8 Ω$
$R_{eff} = 40.8 + \frac{480 \times 20}{480 + 20}$
$= 40.8 + 19.2 = 60 Ω$
$I = \frac{V _{eff}}{R _{eff}} = \frac{45}{60} = 0.75 A$