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  • A galvanometer (coil resistance is converted into an ammeter using a shunt of <img alt="1 \Omega" src="ht

A galvanometer (coil resistance 99 \Omega is converted into an ammeter using a shunt of 1 \Omega and connected as shown in figure -1 . The ammeter reads 3 \mathrm{~A}.

The same galvanometer is converted into a voltmeter by connecting a resistance of 101 \Omega  in series. This voltmeter is connected as shown in figure-2. Its reading is found to be \frac{4}{5} of the full scale reading. The internal resistance of the cell (\mathrm{r}) is
 

Option: 1

1 \Omega


Option: 2

2 \Omega


Option: 3

3 \Omega


Option: 4

4 \Omega


Answers (1)

best_answer

For ammeter

\begin{aligned} & 99 \mathrm{I}_{\mathrm{g}}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) 1 \\ & \text { or } \mathrm{I}=100 \mathrm{I}_{\mathrm{g} \quad \quad \quad \quad \dots(i)} \end{aligned}
\mathrm{I}_{\mathrm{g}} is the full scale deflection current of the galvanometer and \mathrm{I} is the range of ammeter

For the circuit in figure-1

\frac{12 \mathrm{~V}}{2+\mathrm{r}+\frac{49 \times 1}{99+1}}=3 \mathrm{~A} \Rightarrow \mathrm{r}=1.01 \Omega .

Posted by

sudhir kumar

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