Question

The ammeter shown in figure consists of a 480 $\Omega$ coil connected in parallel to a 20 $\Omega$ shunt. The reading of ammeter is

• A. 0.125 A
• B. 1.67 A
• C. 0.13 A
• D. 0.67 A

Answer 1

The correct option is A 0.125 A

The ammeter coil of resistance $480\Omega$ and shunt of $20\Omega$ are connected in parallel, therefore equivalent resistance

$R^{\prime }=\frac{480\times 20}{480+20}=19.2\Omega$

Now, resistance of $140.8\Omega$ and $19.2\Omega$ are in series, hence total resistance in the circuit,

$R=R^{\prime }+19.2=140.8+19.2=160\Omega$

Therefore, current in the circuit (reading by ammeter),

$I=\frac{E}{R}=\frac{20}{160}=0.125A$