Question

To convert a galvanometer into an ammeter a small resistance $S$ (called the shunt) is connected in parallel with a galvanometer. A galvanometer has resistance $G$ and full scale deflection current $i$. To convert this galvanometer into an ammeter of range 10 i, a shunt $S=\frac{G}{9}$ is connected in parallel with G.

Now we want to measure the potential difference with the help of this ammeter. The maximum value of potential difference which can be measured with the help of this is:

• A. $10iG$
• B. $iG$
• C. $9iG$
• D. $\frac{10}{9}iG$

Answer 1

Answer: (B)

$iG$

The voltage across galvanometer is:
$V_g=iG$
When the galvanometer is converted into an ammeter of range $10i$ by applying a shunt resistance $S=\frac{G}{9}$, the current is divided in inverse ratio of their resistance $\frac{1}{9}:1$ between shunt and ammeter .

The current will be in ratio $9:1$.

Current in shunt is $i^{\prime }=\frac{9}{10}10i=9i$

The voltage across shunt resistance is:
$V_s=9i\left(\frac{G}{9}\right)$
Since the galvanometer (ammeter) and shunt are in parallel voltage across them is same, hence the maximum value of potential difference which can be measured with the help of this ammeter is $iG$.