154237
The resistance of an ammeter is $13 \Omega$ and its scale is graduated for a current upto $100 \mathrm{~A}$. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto 750 A by this meter. The value of shunt resistance is
1 $20 \Omega$
2 $2 \Omega$
3 $0.2 \Omega$
4 $2 \mathrm{k} \Omega$
Explanation:
B $\mathrm{R}_{\mathrm{a}}=13 \Omega, \mathrm{I}_{\mathrm{a}}=100 \mathrm{~A}, \mathrm{I}=750 \mathrm{~A}$ From figure, $I_{a} R_{a}=\left(I-I_{a}\right) S$ $100 \times 13=(750-100) \mathrm{S}$ $1300=650 \mathrm{~S}$ $\mathrm{~S}=\frac{1300}{650}=2 \Omega$
UPSEE - 2007
Magnetism and Matter
154239
A moving coil galvanometer has a resistance of $10 \Omega$ and full scale deflection of $0.01 \mathrm{~A}$. It can be converted into voltmeter of $10 \mathrm{~V}$ full scale by connecting into resistance of :
1 $9.90 \Omega$ in series
2 $10 \Omega$ in series
3 $990 \Omega$ in series
4 $0.10 \Omega$ in series
Explanation:
C Given that, $\mathrm{R}_{\mathrm{G}}=10 \Omega, \mathrm{I}_{\mathrm{g}}=0.01 \mathrm{Amp} \text { and } \mathrm{V}_{\text {ext }}=10 \mathrm{~V}$ $\mathrm{~V}_{\mathrm{G}}=\mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{G}}=0.01 \times 10$ $\mathrm{~V}_{\mathrm{G}}=0.1 \mathrm{~V}$ To increase the range of voltmeter resistance $R_{S C}$ is required to connect in series. $\mathrm{R}_{\mathrm{SC}} =\mathrm{R}_{\mathrm{G}}\left(\frac{\mathrm{V}_{\text {ext }}}{\mathrm{V}_{\mathrm{G}}}-1\right)=10 \times\left(\frac{10}{0.1}-1\right)$ $=10(100-1)=10 \times 99$ $\mathrm{R}_{\mathrm{SC}} =990 \Omega$
UPSEE - 2005
Magnetism and Matter
154240
A galvanometer of resistance $G$ is converted into an ammeter using a shunt of resistance $R$. If the ratio of the heat dissipated through the galvanometer and shunt is $3: 4$, then $R$ equals
1 $\frac{4}{3} G$
2 $\frac{3}{4} \mathrm{G}$
3 $\frac{16}{9} \mathrm{G}$
4 $\frac{9}{16} \mathrm{G}$
5 $\mathrm{G}$
Explanation:
B Given, $\frac{\mathrm{H}_{\mathrm{G}}}{\mathrm{H}_{\mathrm{S}}}=\frac{3}{4}, \mathrm{R}_{1}=\mathrm{G}$ We know that, Heat produced in a resistor $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}$ $\therefore \quad \mathrm{H} \propto \frac{1}{\mathrm{R}}$ Hence, ratio of heat dissipated, $\frac{\mathrm{H}_{\mathrm{G}}}{\mathrm{H}_{\mathrm{S}}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ $\frac{3}{4}=\frac{\mathrm{R}_{2}}{\mathrm{G}}$ $\mathrm{R}_{2}=\frac{3}{4} \mathrm{G}$
Kerala CEE - 2016
Magnetism and Matter
154241
The shunt required to send $10 \%$ of the main current through a moving coil galvanometer of resistance $99 \Omega$ is
1 $99 \Omega$
2 $9.9 \Omega$
3 $9 \Omega$
4 $10 \Omega$
5 $11 \Omega$
Explanation:
E Given that, $\mathrm{G}=99 \Omega$ $I_{g}=10 \%$ of $I=\frac{10}{100} I=\frac{I}{10}$ $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{GI}_{\mathrm{g}}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}$ $\mathrm{R}_{\mathrm{S}}=\frac{99 \times \frac{\mathrm{I}}{10}}{\mathrm{I}-\frac{\mathrm{I}}{10}}$ $\mathrm{R}_{\mathrm{S}}=11 \Omega$
154237
The resistance of an ammeter is $13 \Omega$ and its scale is graduated for a current upto $100 \mathrm{~A}$. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto 750 A by this meter. The value of shunt resistance is
1 $20 \Omega$
2 $2 \Omega$
3 $0.2 \Omega$
4 $2 \mathrm{k} \Omega$
Explanation:
B $\mathrm{R}_{\mathrm{a}}=13 \Omega, \mathrm{I}_{\mathrm{a}}=100 \mathrm{~A}, \mathrm{I}=750 \mathrm{~A}$ From figure, $I_{a} R_{a}=\left(I-I_{a}\right) S$ $100 \times 13=(750-100) \mathrm{S}$ $1300=650 \mathrm{~S}$ $\mathrm{~S}=\frac{1300}{650}=2 \Omega$
UPSEE - 2007
Magnetism and Matter
154239
A moving coil galvanometer has a resistance of $10 \Omega$ and full scale deflection of $0.01 \mathrm{~A}$. It can be converted into voltmeter of $10 \mathrm{~V}$ full scale by connecting into resistance of :
1 $9.90 \Omega$ in series
2 $10 \Omega$ in series
3 $990 \Omega$ in series
4 $0.10 \Omega$ in series
Explanation:
C Given that, $\mathrm{R}_{\mathrm{G}}=10 \Omega, \mathrm{I}_{\mathrm{g}}=0.01 \mathrm{Amp} \text { and } \mathrm{V}_{\text {ext }}=10 \mathrm{~V}$ $\mathrm{~V}_{\mathrm{G}}=\mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{G}}=0.01 \times 10$ $\mathrm{~V}_{\mathrm{G}}=0.1 \mathrm{~V}$ To increase the range of voltmeter resistance $R_{S C}$ is required to connect in series. $\mathrm{R}_{\mathrm{SC}} =\mathrm{R}_{\mathrm{G}}\left(\frac{\mathrm{V}_{\text {ext }}}{\mathrm{V}_{\mathrm{G}}}-1\right)=10 \times\left(\frac{10}{0.1}-1\right)$ $=10(100-1)=10 \times 99$ $\mathrm{R}_{\mathrm{SC}} =990 \Omega$
UPSEE - 2005
Magnetism and Matter
154240
A galvanometer of resistance $G$ is converted into an ammeter using a shunt of resistance $R$. If the ratio of the heat dissipated through the galvanometer and shunt is $3: 4$, then $R$ equals
1 $\frac{4}{3} G$
2 $\frac{3}{4} \mathrm{G}$
3 $\frac{16}{9} \mathrm{G}$
4 $\frac{9}{16} \mathrm{G}$
5 $\mathrm{G}$
Explanation:
B Given, $\frac{\mathrm{H}_{\mathrm{G}}}{\mathrm{H}_{\mathrm{S}}}=\frac{3}{4}, \mathrm{R}_{1}=\mathrm{G}$ We know that, Heat produced in a resistor $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}$ $\therefore \quad \mathrm{H} \propto \frac{1}{\mathrm{R}}$ Hence, ratio of heat dissipated, $\frac{\mathrm{H}_{\mathrm{G}}}{\mathrm{H}_{\mathrm{S}}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ $\frac{3}{4}=\frac{\mathrm{R}_{2}}{\mathrm{G}}$ $\mathrm{R}_{2}=\frac{3}{4} \mathrm{G}$
Kerala CEE - 2016
Magnetism and Matter
154241
The shunt required to send $10 \%$ of the main current through a moving coil galvanometer of resistance $99 \Omega$ is
1 $99 \Omega$
2 $9.9 \Omega$
3 $9 \Omega$
4 $10 \Omega$
5 $11 \Omega$
Explanation:
E Given that, $\mathrm{G}=99 \Omega$ $I_{g}=10 \%$ of $I=\frac{10}{100} I=\frac{I}{10}$ $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{GI}_{\mathrm{g}}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}$ $\mathrm{R}_{\mathrm{S}}=\frac{99 \times \frac{\mathrm{I}}{10}}{\mathrm{I}-\frac{\mathrm{I}}{10}}$ $\mathrm{R}_{\mathrm{S}}=11 \Omega$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Magnetism and Matter
154237
The resistance of an ammeter is $13 \Omega$ and its scale is graduated for a current upto $100 \mathrm{~A}$. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto 750 A by this meter. The value of shunt resistance is
1 $20 \Omega$
2 $2 \Omega$
3 $0.2 \Omega$
4 $2 \mathrm{k} \Omega$
Explanation:
B $\mathrm{R}_{\mathrm{a}}=13 \Omega, \mathrm{I}_{\mathrm{a}}=100 \mathrm{~A}, \mathrm{I}=750 \mathrm{~A}$ From figure, $I_{a} R_{a}=\left(I-I_{a}\right) S$ $100 \times 13=(750-100) \mathrm{S}$ $1300=650 \mathrm{~S}$ $\mathrm{~S}=\frac{1300}{650}=2 \Omega$
UPSEE - 2007
Magnetism and Matter
154239
A moving coil galvanometer has a resistance of $10 \Omega$ and full scale deflection of $0.01 \mathrm{~A}$. It can be converted into voltmeter of $10 \mathrm{~V}$ full scale by connecting into resistance of :
1 $9.90 \Omega$ in series
2 $10 \Omega$ in series
3 $990 \Omega$ in series
4 $0.10 \Omega$ in series
Explanation:
C Given that, $\mathrm{R}_{\mathrm{G}}=10 \Omega, \mathrm{I}_{\mathrm{g}}=0.01 \mathrm{Amp} \text { and } \mathrm{V}_{\text {ext }}=10 \mathrm{~V}$ $\mathrm{~V}_{\mathrm{G}}=\mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{G}}=0.01 \times 10$ $\mathrm{~V}_{\mathrm{G}}=0.1 \mathrm{~V}$ To increase the range of voltmeter resistance $R_{S C}$ is required to connect in series. $\mathrm{R}_{\mathrm{SC}} =\mathrm{R}_{\mathrm{G}}\left(\frac{\mathrm{V}_{\text {ext }}}{\mathrm{V}_{\mathrm{G}}}-1\right)=10 \times\left(\frac{10}{0.1}-1\right)$ $=10(100-1)=10 \times 99$ $\mathrm{R}_{\mathrm{SC}} =990 \Omega$
UPSEE - 2005
Magnetism and Matter
154240
A galvanometer of resistance $G$ is converted into an ammeter using a shunt of resistance $R$. If the ratio of the heat dissipated through the galvanometer and shunt is $3: 4$, then $R$ equals
1 $\frac{4}{3} G$
2 $\frac{3}{4} \mathrm{G}$
3 $\frac{16}{9} \mathrm{G}$
4 $\frac{9}{16} \mathrm{G}$
5 $\mathrm{G}$
Explanation:
B Given, $\frac{\mathrm{H}_{\mathrm{G}}}{\mathrm{H}_{\mathrm{S}}}=\frac{3}{4}, \mathrm{R}_{1}=\mathrm{G}$ We know that, Heat produced in a resistor $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}$ $\therefore \quad \mathrm{H} \propto \frac{1}{\mathrm{R}}$ Hence, ratio of heat dissipated, $\frac{\mathrm{H}_{\mathrm{G}}}{\mathrm{H}_{\mathrm{S}}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ $\frac{3}{4}=\frac{\mathrm{R}_{2}}{\mathrm{G}}$ $\mathrm{R}_{2}=\frac{3}{4} \mathrm{G}$
Kerala CEE - 2016
Magnetism and Matter
154241
The shunt required to send $10 \%$ of the main current through a moving coil galvanometer of resistance $99 \Omega$ is
1 $99 \Omega$
2 $9.9 \Omega$
3 $9 \Omega$
4 $10 \Omega$
5 $11 \Omega$
Explanation:
E Given that, $\mathrm{G}=99 \Omega$ $I_{g}=10 \%$ of $I=\frac{10}{100} I=\frac{I}{10}$ $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{GI}_{\mathrm{g}}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}$ $\mathrm{R}_{\mathrm{S}}=\frac{99 \times \frac{\mathrm{I}}{10}}{\mathrm{I}-\frac{\mathrm{I}}{10}}$ $\mathrm{R}_{\mathrm{S}}=11 \Omega$
154237
The resistance of an ammeter is $13 \Omega$ and its scale is graduated for a current upto $100 \mathrm{~A}$. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto 750 A by this meter. The value of shunt resistance is
1 $20 \Omega$
2 $2 \Omega$
3 $0.2 \Omega$
4 $2 \mathrm{k} \Omega$
Explanation:
B $\mathrm{R}_{\mathrm{a}}=13 \Omega, \mathrm{I}_{\mathrm{a}}=100 \mathrm{~A}, \mathrm{I}=750 \mathrm{~A}$ From figure, $I_{a} R_{a}=\left(I-I_{a}\right) S$ $100 \times 13=(750-100) \mathrm{S}$ $1300=650 \mathrm{~S}$ $\mathrm{~S}=\frac{1300}{650}=2 \Omega$
UPSEE - 2007
Magnetism and Matter
154239
A moving coil galvanometer has a resistance of $10 \Omega$ and full scale deflection of $0.01 \mathrm{~A}$. It can be converted into voltmeter of $10 \mathrm{~V}$ full scale by connecting into resistance of :
1 $9.90 \Omega$ in series
2 $10 \Omega$ in series
3 $990 \Omega$ in series
4 $0.10 \Omega$ in series
Explanation:
C Given that, $\mathrm{R}_{\mathrm{G}}=10 \Omega, \mathrm{I}_{\mathrm{g}}=0.01 \mathrm{Amp} \text { and } \mathrm{V}_{\text {ext }}=10 \mathrm{~V}$ $\mathrm{~V}_{\mathrm{G}}=\mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{G}}=0.01 \times 10$ $\mathrm{~V}_{\mathrm{G}}=0.1 \mathrm{~V}$ To increase the range of voltmeter resistance $R_{S C}$ is required to connect in series. $\mathrm{R}_{\mathrm{SC}} =\mathrm{R}_{\mathrm{G}}\left(\frac{\mathrm{V}_{\text {ext }}}{\mathrm{V}_{\mathrm{G}}}-1\right)=10 \times\left(\frac{10}{0.1}-1\right)$ $=10(100-1)=10 \times 99$ $\mathrm{R}_{\mathrm{SC}} =990 \Omega$
UPSEE - 2005
Magnetism and Matter
154240
A galvanometer of resistance $G$ is converted into an ammeter using a shunt of resistance $R$. If the ratio of the heat dissipated through the galvanometer and shunt is $3: 4$, then $R$ equals
1 $\frac{4}{3} G$
2 $\frac{3}{4} \mathrm{G}$
3 $\frac{16}{9} \mathrm{G}$
4 $\frac{9}{16} \mathrm{G}$
5 $\mathrm{G}$
Explanation:
B Given, $\frac{\mathrm{H}_{\mathrm{G}}}{\mathrm{H}_{\mathrm{S}}}=\frac{3}{4}, \mathrm{R}_{1}=\mathrm{G}$ We know that, Heat produced in a resistor $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}$ $\therefore \quad \mathrm{H} \propto \frac{1}{\mathrm{R}}$ Hence, ratio of heat dissipated, $\frac{\mathrm{H}_{\mathrm{G}}}{\mathrm{H}_{\mathrm{S}}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ $\frac{3}{4}=\frac{\mathrm{R}_{2}}{\mathrm{G}}$ $\mathrm{R}_{2}=\frac{3}{4} \mathrm{G}$
Kerala CEE - 2016
Magnetism and Matter
154241
The shunt required to send $10 \%$ of the main current through a moving coil galvanometer of resistance $99 \Omega$ is
1 $99 \Omega$
2 $9.9 \Omega$
3 $9 \Omega$
4 $10 \Omega$
5 $11 \Omega$
Explanation:
E Given that, $\mathrm{G}=99 \Omega$ $I_{g}=10 \%$ of $I=\frac{10}{100} I=\frac{I}{10}$ $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{GI}_{\mathrm{g}}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}$ $\mathrm{R}_{\mathrm{S}}=\frac{99 \times \frac{\mathrm{I}}{10}}{\mathrm{I}-\frac{\mathrm{I}}{10}}$ $\mathrm{R}_{\mathrm{S}}=11 \Omega$
