Question

An ammeter of resistance $480\Omega$ with shunt resistance $20\Omega$ is connected in series with a resistance $140.8\Omega$. A 20V supply is given to the circuit, what will be the reading of the ammeter?
A. 1.25A
B. 0.125A
C. 12.5A
D. 0.0125A

Answer 1

A galvanometer is converted to an ammeter by connecting a shunt resistance parallel with the galvanometer. This means resistance $480\Omega$ is connected parallel to $20\Omega$ to result in a low resistance. Find the resultant resistance, say R. R is then connected in series with a resistance $140.8\Omega$. Find the resultant resistance, say RE. Substitute the values of obtained resistance RE and voltage given to calculate the current passing through the ammeter.

Complete step by step answer:
We are given that an ammeter of resistance $480\Omega$ with shunt resistance $20\Omega$ is connected in series with a resistance $140.8\Omega$ and a 20V supply is given to the circuit.
We have to calculate the reading (current) off the ammeter.
Resistance $480\Omega$ is connected with a shunt resistance $20\Omega$, this means shunt resistance is connected in parallel to the galvanometer to keep its resistance low. Low resistance galvanometer is also called an ammeter.
When two resistances are connected in parallel, then the resultant resistance (R) will be
$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} \\\ {R_1} = 480,{R_2} = 20 \\\ \implies \dfrac{1}{R} = \dfrac{1}{{480}} + \dfrac{1}{{20}} \\\ \implies \dfrac{1}{R} = \dfrac{{1 + 24}}{{480}} = \dfrac{{25}}{{480}} \\\ \implies R = \dfrac{{480}}{{25}} \\\ R = 19.2\Omega \\\$
And the resultant resistance $19.2\Omega$ is connected in series with $140.8\Omega$
When two resistances are connected in series, then the resultant resistance (R) will be
$R = {R_1} + {R_2} \\\ \implies {R_1} = 19.2,{R_2} = 140.8 \\\ \implies R = 19.2 + 140.8 \\\ \implies R = 160\Omega \\\$
The final resultant resistance of the circuit is $160\Omega$
The reading (current) off the ammeter will be
$I = \dfrac{V}{R}$, where V is the Voltage supply given to the circuit and R is the resistance.
$I = \dfrac{V}{R} \\\ V = 20V,R = 160\Omega \\\ I = \dfrac{{20}}{{160}} \\\ \implies I = \dfrac{1}{8} \\\ \therefore I = 0.125A \\\$
The reading off the Ammeter will be 0.125 Amperes.

So, the correct answer is “Option B”.

Note:
A shunt is a device which creates a low resistance path for electric current in a circuit. Ammeter is a type of galvanometer. An ammeter just shows the magnitude of the current whereas a galvanometer shows both the magnitude and the direction of the current. We do not use a galvanometer as an ammeter because the galvanometer shows very large deflection for a very small current whereas an ammeter does not.