Question
A circuit contains an ammeter, a battery of $$30\,V$$ and a resistance $$40.8\,\Omega $$ all connected in series. If the ammeter has a coil of resistance $$480\,\Omega $$ and a shunt of $$20\,\Omega ,$$ then reading in the ammeter will be
A.
$$0.5\,A$$
B.
$$0.25\,A$$
C.
$$2\,A$$
D.
$$1\,A$$
Answer :
$$0.5\,A$$
Solution :
Effective resistance of a circuit,
$${R_{{\text{eff}}}} = 40.8 + \frac{{480 \times 20}}{{480 + 20}} = 40.8 + 19.2 = 60\,\Omega $$
So, current flowing across ammeter,
$$I = \frac{V}{R} = \frac{{30}}{{60}} = \frac{1}{2} = 0.5\,A$$
Hence, reading of ammeter $$= 0.5\,A$$
Effective resistance of a circuit,
$${R_{{\text{eff}}}} = 40.8 + \frac{{480 \times 20}}{{480 + 20}} = 40.8 + 19.2 = 60\,\Omega $$
So, current flowing across ammeter,
$$I = \frac{V}{R} = \frac{{30}}{{60}} = \frac{1}{2} = 0.5\,A$$
Hence, reading of ammeter $$= 0.5\,A$$
