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Magnetism and Matter

154237 The resistance of an ammeter is $13 Ω$ and its scale is graduated for a current upto $100 A$ . 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 Ω

2

2 Ω

3

0.2 Ω

4

2 k Ω

Explanation:

B $R_{a} = 13 Ω, I_{a} = 100 A, I = 750 A$

From figure, $I_{a} R_{a} = (I - I_{a}) S$
$100 \times 13 = (750 - 100) S$
$1300 = 650 S$
$S = \frac{1300}{650} = 2 Ω$

UPSEE - 2007

Magnetism and Matter

154239 A moving coil galvanometer has a resistance of $10 Ω$ and full scale deflection of $0.01 A$ . It can be converted into voltmeter of $10 V$ full scale by connecting into resistance of :

1

9.90 Ω

in series

2

10 Ω

in series

3

990 Ω

in series

4

0.10 Ω

in series

Explanation:

C Given that,
$R_{G} = 10 Ω, I_{g} = 0.01 Amp and V_{ext} = 10 V$
$V_{G} = I_{g} R_{G} = 0.01 \times 10$
$V_{G} = 0.1 V$
To increase the range of voltmeter resistance $R_{S C}$ is required to connect in series.
$R_{SC} = R_{G} (\frac{V_{ext}}{V_{G}} - 1) = 10 \times (\frac{10}{0.1} - 1)$
$= 10 (100 - 1) = 10 \times 99$
$R_{SC} = 990 Ω$

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} G

3

\frac{16}{9} G

4

\frac{9}{16} G

5

G

Explanation:

B Given,
$\frac{H_{G}}{H_{S}} = \frac{3}{4}, R_{1} = G$
We know that,
Heat produced in a resistor $H = I^{2} Rt = \frac{V^{2}}{R} t$
$∴ H \propto \frac{1}{R}$
Hence, ratio of heat dissipated,
$\frac{H_{G}}{H_{S}} = \frac{R_{2}}{R_{1}}$
$\frac{3}{4} = \frac{R_{2}}{G}$
$R_{2} = \frac{3}{4} G$

Kerala CEE - 2016

Magnetism and Matter

154241 The shunt required to send $10 %$ of the main current through a moving coil galvanometer of resistance $99 Ω$ is

1

99 Ω

2

9.9 Ω

3

9 Ω

4

10 Ω

5

11 Ω

Explanation:

E Given that,
$G = 99 Ω$
$I_{g} = 10 %$ of $I = \frac{10}{100} I = \frac{I}{10}$
$R_{S} = \frac{{GI}_{g}}{I - I_{g}}$
$R_{S} = \frac{99 \times \frac{I}{10}}{I - \frac{I}{10}}$
$R_{S} = 11 Ω$

Kerala CEE - 2015

Magnetism and Matter

154237 The resistance of an ammeter is $13 Ω$ and its scale is graduated for a current upto $100 A$ . 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 Ω

2

2 Ω

3

0.2 Ω

4

2 k Ω

Explanation:

B $R_{a} = 13 Ω, I_{a} = 100 A, I = 750 A$

From figure, $I_{a} R_{a} = (I - I_{a}) S$
$100 \times 13 = (750 - 100) S$
$1300 = 650 S$
$S = \frac{1300}{650} = 2 Ω$

UPSEE - 2007

Magnetism and Matter

154239 A moving coil galvanometer has a resistance of $10 Ω$ and full scale deflection of $0.01 A$ . It can be converted into voltmeter of $10 V$ full scale by connecting into resistance of :

1

9.90 Ω

in series

2

10 Ω

in series

3

990 Ω

in series

4

0.10 Ω

in series

Explanation:

C Given that,
$R_{G} = 10 Ω, I_{g} = 0.01 Amp and V_{ext} = 10 V$
$V_{G} = I_{g} R_{G} = 0.01 \times 10$
$V_{G} = 0.1 V$
To increase the range of voltmeter resistance $R_{S C}$ is required to connect in series.
$R_{SC} = R_{G} (\frac{V_{ext}}{V_{G}} - 1) = 10 \times (\frac{10}{0.1} - 1)$
$= 10 (100 - 1) = 10 \times 99$
$R_{SC} = 990 Ω$

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} G

3

\frac{16}{9} G

4

\frac{9}{16} G

5

G

Explanation:

B Given,
$\frac{H_{G}}{H_{S}} = \frac{3}{4}, R_{1} = G$
We know that,
Heat produced in a resistor $H = I^{2} Rt = \frac{V^{2}}{R} t$
$∴ H \propto \frac{1}{R}$
Hence, ratio of heat dissipated,
$\frac{H_{G}}{H_{S}} = \frac{R_{2}}{R_{1}}$
$\frac{3}{4} = \frac{R_{2}}{G}$
$R_{2} = \frac{3}{4} G$

Kerala CEE - 2016

Magnetism and Matter

154241 The shunt required to send $10 %$ of the main current through a moving coil galvanometer of resistance $99 Ω$ is

1

99 Ω

2

9.9 Ω

3

9 Ω

4

10 Ω

5

11 Ω

Explanation:

E Given that,
$G = 99 Ω$
$I_{g} = 10 %$ of $I = \frac{10}{100} I = \frac{I}{10}$
$R_{S} = \frac{{GI}_{g}}{I - I_{g}}$
$R_{S} = \frac{99 \times \frac{I}{10}}{I - \frac{I}{10}}$
$R_{S} = 11 Ω$

Kerala CEE - 2015

Magnetism and Matter

154237 The resistance of an ammeter is $13 Ω$ and its scale is graduated for a current upto $100 A$ . 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 Ω

2

2 Ω

3

0.2 Ω

4

2 k Ω

Explanation:

B $R_{a} = 13 Ω, I_{a} = 100 A, I = 750 A$

From figure, $I_{a} R_{a} = (I - I_{a}) S$
$100 \times 13 = (750 - 100) S$
$1300 = 650 S$
$S = \frac{1300}{650} = 2 Ω$

UPSEE - 2007

Magnetism and Matter

154239 A moving coil galvanometer has a resistance of $10 Ω$ and full scale deflection of $0.01 A$ . It can be converted into voltmeter of $10 V$ full scale by connecting into resistance of :

1

9.90 Ω

in series

2

10 Ω

in series

3

990 Ω

in series

4

0.10 Ω

in series

Explanation:

C Given that,
$R_{G} = 10 Ω, I_{g} = 0.01 Amp and V_{ext} = 10 V$
$V_{G} = I_{g} R_{G} = 0.01 \times 10$
$V_{G} = 0.1 V$
To increase the range of voltmeter resistance $R_{S C}$ is required to connect in series.
$R_{SC} = R_{G} (\frac{V_{ext}}{V_{G}} - 1) = 10 \times (\frac{10}{0.1} - 1)$
$= 10 (100 - 1) = 10 \times 99$
$R_{SC} = 990 Ω$

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} G

3

\frac{16}{9} G

4

\frac{9}{16} G

5

G

Explanation:

B Given,
$\frac{H_{G}}{H_{S}} = \frac{3}{4}, R_{1} = G$
We know that,
Heat produced in a resistor $H = I^{2} Rt = \frac{V^{2}}{R} t$
$∴ H \propto \frac{1}{R}$
Hence, ratio of heat dissipated,
$\frac{H_{G}}{H_{S}} = \frac{R_{2}}{R_{1}}$
$\frac{3}{4} = \frac{R_{2}}{G}$
$R_{2} = \frac{3}{4} G$

Kerala CEE - 2016

Magnetism and Matter

154241 The shunt required to send $10 %$ of the main current through a moving coil galvanometer of resistance $99 Ω$ is

1

99 Ω

2

9.9 Ω

3

9 Ω

4

10 Ω

5

11 Ω

Explanation:

E Given that,
$G = 99 Ω$
$I_{g} = 10 %$ of $I = \frac{10}{100} I = \frac{I}{10}$
$R_{S} = \frac{{GI}_{g}}{I - I_{g}}$
$R_{S} = \frac{99 \times \frac{I}{10}}{I - \frac{I}{10}}$
$R_{S} = 11 Ω$

Kerala CEE - 2015

Magnetism and Matter

154237 The resistance of an ammeter is $13 Ω$ and its scale is graduated for a current upto $100 A$ . 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 Ω

2

2 Ω

3

0.2 Ω

4

2 k Ω

Explanation:

B $R_{a} = 13 Ω, I_{a} = 100 A, I = 750 A$

From figure, $I_{a} R_{a} = (I - I_{a}) S$
$100 \times 13 = (750 - 100) S$
$1300 = 650 S$
$S = \frac{1300}{650} = 2 Ω$

UPSEE - 2007

Magnetism and Matter

154239 A moving coil galvanometer has a resistance of $10 Ω$ and full scale deflection of $0.01 A$ . It can be converted into voltmeter of $10 V$ full scale by connecting into resistance of :

1

9.90 Ω

in series

2

10 Ω

in series

3

990 Ω

in series

4

0.10 Ω

in series

Explanation:

C Given that,
$R_{G} = 10 Ω, I_{g} = 0.01 Amp and V_{ext} = 10 V$
$V_{G} = I_{g} R_{G} = 0.01 \times 10$
$V_{G} = 0.1 V$
To increase the range of voltmeter resistance $R_{S C}$ is required to connect in series.
$R_{SC} = R_{G} (\frac{V_{ext}}{V_{G}} - 1) = 10 \times (\frac{10}{0.1} - 1)$
$= 10 (100 - 1) = 10 \times 99$
$R_{SC} = 990 Ω$

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} G

3

\frac{16}{9} G

4

\frac{9}{16} G

5

G

Explanation:

B Given,
$\frac{H_{G}}{H_{S}} = \frac{3}{4}, R_{1} = G$
We know that,
Heat produced in a resistor $H = I^{2} Rt = \frac{V^{2}}{R} t$
$∴ H \propto \frac{1}{R}$
Hence, ratio of heat dissipated,
$\frac{H_{G}}{H_{S}} = \frac{R_{2}}{R_{1}}$
$\frac{3}{4} = \frac{R_{2}}{G}$
$R_{2} = \frac{3}{4} G$

Kerala CEE - 2016

Magnetism and Matter

154241 The shunt required to send $10 %$ of the main current through a moving coil galvanometer of resistance $99 Ω$ is

1

99 Ω

2

9.9 Ω

3

9 Ω

4

10 Ω

5

11 Ω

Explanation:

E Given that,
$G = 99 Ω$
$I_{g} = 10 %$ of $I = \frac{10}{100} I = \frac{I}{10}$
$R_{S} = \frac{{GI}_{g}}{I - I_{g}}$
$R_{S} = \frac{99 \times \frac{I}{10}}{I - \frac{I}{10}}$
$R_{S} = 11 Ω$

Kerala CEE - 2015