Question

# A circuit contains an ammeter, a battery of 30V and a resistance 40.8ohm all connected in series. If the ammeter has a coil of resistance 480ohm and a shunt of 20ohm, the reading in the ammeter will be:(A). 1A(B). 0.5A(C). 0.25A(D). 2A

Hint: The ammeter is in series with the 40.8 ohm resistance, and ammeter is in parallel to the shunt resistance. We know that ohm’s law states that voltage or potential difference in a circuit is completely proportional to the current and resistance of the circuit.

Complete step-by-step solution -

Now, According to the problem
If, we consider resistance of ammeter to be ${{R}_{1}}$ and shunt resistance as ${{R}_{2}}$.
And both of them are in parallel to each other then we can find the equivalent resistance of that circuit.
${{R}_{eq}}=\dfrac{{{R}_{1}}\times {{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}$
${{R}_{eq}}=\dfrac{480\times 20}{480+20}$
${{R}_{eq}}=\dfrac{9600}{500}$
${{R}_{eq}}=19.2\Omega$
Now the circuit looks like,

Now let us consider ,
40.8 ohm as R$_{3}$,
R$_{3}$ and Req are in series with each other,
Therefore,
${{R}_{eq1}}={{R}_{3}}+{{R}_{eq}}$
${{R}_{eq1}}=$ 40.8+19.2 ohm.
${{R}_{eq1}}$ =60ohm.
Now from ohm law, V=IR;
V=30V
R=60ohm
I=V/R
I=30/60
I=1/2 or 0.5A.
Therefore option B (0.5A) is the correct option.