JEE MAIN - Physics (2023 - 15th April Morning Shift - No. 13)
For designing a voltmeter of range $50 \mathrm{~V}$ and an ammeter of range $10 \mathrm{~mA}$ using a galvanometer which has a coil of resistance $54 \Omega$ showing a full scale deflection for $1 \mathrm{~mA}$ as in figure.
_15th_April_Morning_Shift_en_13_1.png)
(A) for voltmeter $R \approx 50 \mathrm{k} \Omega$
(B) for ammeter $\mathrm{r} \approx 0.2 \Omega$
(C) for ammeter $\mathrm{r}=6 \Omega$
(D) for voltmeter $R \approx 5 \mathrm{k} \Omega$
(E) for voltmeter $R \approx 500 \Omega$
Choose the correct answer from the options given below:
(A) for voltmeter $R \approx 50 \mathrm{k} \Omega$
(B) for ammeter $\mathrm{r} \approx 0.2 \Omega$
(C) for ammeter $\mathrm{r}=6 \Omega$
(D) for voltmeter $R \approx 5 \mathrm{k} \Omega$
(E) for voltmeter $R \approx 500 \Omega$
Choose the correct answer from the options given below:
$(\mathrm{A})$ and $(\mathrm{C})$
(A) and (B)
(C) and (E)
(C) and (D)
Explanation
For voltmeter
$$ \begin{aligned} & \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G} \\\\ & =\frac{50}{10^{-3}}-54 \approx 50 \mathrm{k} \Omega~(\mathrm{A}) \end{aligned} $$
For ammeter
$$ \mathrm{S}=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}=\frac{10^{-3} \times 54}{(10-1) \times 10^{-3}}=6 \Omega~(\mathrm{C}) $$
$$ \begin{aligned} & \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G} \\\\ & =\frac{50}{10^{-3}}-54 \approx 50 \mathrm{k} \Omega~(\mathrm{A}) \end{aligned} $$
For ammeter
$$ \mathrm{S}=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}=\frac{10^{-3} \times 54}{(10-1) \times 10^{-3}}=6 \Omega~(\mathrm{C}) $$
Comments (0)
