Question

An ammeter of resistance $480\Omega$ with shunt resistance $20\Omega$ is connected in series with a reasistance $140.8\Omega$.A $20V$ supply is given to circuit,What will be the reading off ammeter?

• A. $1.25A$
• B. $0.125A$
• C. $12.5A$
• D. $0.0125A$

Answer 1

The correct option is B $0.125A$

Ammeter resistance $=480$ ohm
Shunt resistance $=20$ ohm
In order to convert a Galvanometer into an Ammeter, "shunt" resistance of very low magnitude of resistance needs to be connected in parallel to Galvanometer.
So when ammeter is shunted,
Resistance $R_1$ :
$R_1$=($\frac{1}{480}$+$\frac{1}{20}$)${}^{-}1$=19.2 ohm
Ammeter connected in series with resistance $140.8$ ohm
So equivalent resistance R:
$R=R_1+140.8=160$ ohm
A $20V$ supply is given to circuit.

$\therefore$ reading off ammeter $I$:
$I=20/R=20/160=0.125A$