QBManager

Magnetism and Matter

154246 The radius of the coil of a TG which has 10 turns is $0.1 m$ . The current required to produce a deflection of $60^{\circ} (B_{h} = 4 \times 10^{- 5} T)$ is:

1

3 A

2

1.1 A

3

2.1 A

4

1.5 A

5

2.6 A

Explanation:

B Given, radius of coil $= 0.1 m, θ = 60^{\circ}$
$B = B_{H} \tan θ$
$\frac{μ_{o} ni}{2 r} = B_{H} \tan θ$
$i = \frac{2 {rB}_{H} \tan θ}{μ_{o} n} = \frac{2 \times 0.1 \times 4 \times 10^{- 5} \times \tan 60^{\circ}}{4 π \times 10^{- 7} \times 10}$
$i = \frac{2 \times 0.1 \times 4 \times 10^{- 5} \times \sqrt{3}}{10 \times 4 π \times 10^{- 7}}$
$i = 1.1 Amp$

Kerala CEE 2004

Magnetism and Matter

154247 A $50 Ω$ galvanometer is shunted by a resistance of $5 Ω$ . The percentage of the total current, which passes through the galvanometer is

1

8.1 %

2

10.1 %

3

11.1 %

4

9.1 %

Explanation:

D Given, $R_{G} = 50 Ω, R_{S} = 5 Ω$
$I_{G} \times R_{G} = (I - I_{G}) R_{S}$
$I_{G} \times 50 = (I - I_{G}) 5$
$10 I_{G} = I - I_{G}$
$I = 11 I_{G}$
$I_{G} = \frac{I}{11}$
Let $%$ of current passing through galvanometer is $x$ $x %$ of $I = I_{G}$
$\frac{x}{100} \times I = \frac{I}{11}$
$x = \frac{100}{11} = 9.09 = 9.1 %$

JCECE-2015

Magnetism and Matter

154248 A galvanometer coil has a resistance of $15 Ω$ and gives full scale deflection for a current of 4 $m A$ . To convert it to an ammeter of range 0 to $6 A$

1

10 m Ω

resistance is to be connected in parallel to the galvanometer

2

10 m Ω

resistance is to be connected in series with the galvanometer

3

0.1 Ω

resistance is to be connected in parallel to the galvanometer

4

0.1 Ω

resistance is to be connected in series with the galvanometer

Explanation:

A $G = 15 Ω$
$I_{g} = 4 mA = 4 \times 10^{- 3} Amp$
Total current $(I) = 6 A$
Formula used,
$S = \frac{I_{g} G}{I - I_{g}}$
$S = \frac{4 \times 10^{- 3} \times 15}{6 - 4 \times 10^{- 3}}$
$S = \frac{4 \times 10^{- 3} \times 15 \times 10^{3}}{6000 - 4}$
$S = \frac{60}{5996}$
$S = 0.0100$
$S = 10 \times 10^{- 3} Ω$
$S = 10 m Ω$
Therefore a resistance of $10 m Ω$ should be connected in parallel to the galvanometer.

JCECE-2009

Magnetism and Matter

154249 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 Resistance of an ammeter $(R_{A}) = 13 Ω$
Current flowing through $R_{A}$ is $I_{A} = 100 A$
Let Shunt resistance be $R_{S}$
$∴$ Current flowing through $R_{S}$ ,
$(I_{S}) = 750 - I_{A}$
$I_{S} = 750 - 100$
$I_{S} = 650 A$
Potential difference across ammeter and shunt is same
$I_{S} \cdot R_{S} = I_{A} \cdot R_{A}$
$650 \times R_{S} = 100 \times 13$
$R_{S} = 2 Ω$

JCECE-2008

Magnetism and Matter

154250 In an ammeter, $4 %$ of the main current is passing through galvanometer. If the galvanometer is shunted with a $5 Ω$ resistance, the resistance of the galvanometer is :

1

120 Ω

2

20 Ω

3

5 Ω

4

4 Ω

Explanation:

A Given,
$I_{G} = 4 % of I, I_{G} = \frac{4}{100} I$
$\frac{I_{G}}{I} = \frac{4}{100}$
$I_{G} \times R_{G} = I_{S} \times R_{S}$
$I_{G} \times R_{G} = (I - I_{G}) \times R_{S}$
$R_{G} = (\frac{I}{I_{G}} - 1) R_{S}$
$∴ R_{G} = (\frac{100}{4} - 1) 5 = (25 - 1) 5 = 120 Ω$

JCECE-2004

Magnetism and Matter

154246 The radius of the coil of a TG which has 10 turns is $0.1 m$ . The current required to produce a deflection of $60^{\circ} (B_{h} = 4 \times 10^{- 5} T)$ is:

1

3 A

2

1.1 A

3

2.1 A

4

1.5 A

5

2.6 A

Explanation:

B Given, radius of coil $= 0.1 m, θ = 60^{\circ}$
$B = B_{H} \tan θ$
$\frac{μ_{o} ni}{2 r} = B_{H} \tan θ$
$i = \frac{2 {rB}_{H} \tan θ}{μ_{o} n} = \frac{2 \times 0.1 \times 4 \times 10^{- 5} \times \tan 60^{\circ}}{4 π \times 10^{- 7} \times 10}$
$i = \frac{2 \times 0.1 \times 4 \times 10^{- 5} \times \sqrt{3}}{10 \times 4 π \times 10^{- 7}}$
$i = 1.1 Amp$

Kerala CEE 2004

Magnetism and Matter

154247 A $50 Ω$ galvanometer is shunted by a resistance of $5 Ω$ . The percentage of the total current, which passes through the galvanometer is

1

8.1 %

2

10.1 %

3

11.1 %

4

9.1 %

Explanation:

D Given, $R_{G} = 50 Ω, R_{S} = 5 Ω$
$I_{G} \times R_{G} = (I - I_{G}) R_{S}$
$I_{G} \times 50 = (I - I_{G}) 5$
$10 I_{G} = I - I_{G}$
$I = 11 I_{G}$
$I_{G} = \frac{I}{11}$
Let $%$ of current passing through galvanometer is $x$ $x %$ of $I = I_{G}$
$\frac{x}{100} \times I = \frac{I}{11}$
$x = \frac{100}{11} = 9.09 = 9.1 %$

JCECE-2015

Magnetism and Matter

154248 A galvanometer coil has a resistance of $15 Ω$ and gives full scale deflection for a current of 4 $m A$ . To convert it to an ammeter of range 0 to $6 A$

1

10 m Ω

resistance is to be connected in parallel to the galvanometer

2

10 m Ω

resistance is to be connected in series with the galvanometer

3

0.1 Ω

resistance is to be connected in parallel to the galvanometer

4

0.1 Ω

resistance is to be connected in series with the galvanometer

Explanation:

A $G = 15 Ω$
$I_{g} = 4 mA = 4 \times 10^{- 3} Amp$
Total current $(I) = 6 A$
Formula used,
$S = \frac{I_{g} G}{I - I_{g}}$
$S = \frac{4 \times 10^{- 3} \times 15}{6 - 4 \times 10^{- 3}}$
$S = \frac{4 \times 10^{- 3} \times 15 \times 10^{3}}{6000 - 4}$
$S = \frac{60}{5996}$
$S = 0.0100$
$S = 10 \times 10^{- 3} Ω$
$S = 10 m Ω$
Therefore a resistance of $10 m Ω$ should be connected in parallel to the galvanometer.

JCECE-2009

Magnetism and Matter

154249 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 Resistance of an ammeter $(R_{A}) = 13 Ω$
Current flowing through $R_{A}$ is $I_{A} = 100 A$
Let Shunt resistance be $R_{S}$
$∴$ Current flowing through $R_{S}$ ,
$(I_{S}) = 750 - I_{A}$
$I_{S} = 750 - 100$
$I_{S} = 650 A$
Potential difference across ammeter and shunt is same
$I_{S} \cdot R_{S} = I_{A} \cdot R_{A}$
$650 \times R_{S} = 100 \times 13$
$R_{S} = 2 Ω$

JCECE-2008

Magnetism and Matter

154250 In an ammeter, $4 %$ of the main current is passing through galvanometer. If the galvanometer is shunted with a $5 Ω$ resistance, the resistance of the galvanometer is :

1

120 Ω

2

20 Ω

3

5 Ω

4

4 Ω

Explanation:

A Given,
$I_{G} = 4 % of I, I_{G} = \frac{4}{100} I$
$\frac{I_{G}}{I} = \frac{4}{100}$
$I_{G} \times R_{G} = I_{S} \times R_{S}$
$I_{G} \times R_{G} = (I - I_{G}) \times R_{S}$
$R_{G} = (\frac{I}{I_{G}} - 1) R_{S}$
$∴ R_{G} = (\frac{100}{4} - 1) 5 = (25 - 1) 5 = 120 Ω$

JCECE-2004

Magnetism and Matter

154246 The radius of the coil of a TG which has 10 turns is $0.1 m$ . The current required to produce a deflection of $60^{\circ} (B_{h} = 4 \times 10^{- 5} T)$ is:

1

3 A

2

1.1 A

3

2.1 A

4

1.5 A

5

2.6 A

Explanation:

B Given, radius of coil $= 0.1 m, θ = 60^{\circ}$
$B = B_{H} \tan θ$
$\frac{μ_{o} ni}{2 r} = B_{H} \tan θ$
$i = \frac{2 {rB}_{H} \tan θ}{μ_{o} n} = \frac{2 \times 0.1 \times 4 \times 10^{- 5} \times \tan 60^{\circ}}{4 π \times 10^{- 7} \times 10}$
$i = \frac{2 \times 0.1 \times 4 \times 10^{- 5} \times \sqrt{3}}{10 \times 4 π \times 10^{- 7}}$
$i = 1.1 Amp$

Kerala CEE 2004

Magnetism and Matter

154247 A $50 Ω$ galvanometer is shunted by a resistance of $5 Ω$ . The percentage of the total current, which passes through the galvanometer is

1

8.1 %

2

10.1 %

3

11.1 %

4

9.1 %

Explanation:

D Given, $R_{G} = 50 Ω, R_{S} = 5 Ω$
$I_{G} \times R_{G} = (I - I_{G}) R_{S}$
$I_{G} \times 50 = (I - I_{G}) 5$
$10 I_{G} = I - I_{G}$
$I = 11 I_{G}$
$I_{G} = \frac{I}{11}$
Let $%$ of current passing through galvanometer is $x$ $x %$ of $I = I_{G}$
$\frac{x}{100} \times I = \frac{I}{11}$
$x = \frac{100}{11} = 9.09 = 9.1 %$

JCECE-2015

Magnetism and Matter

154248 A galvanometer coil has a resistance of $15 Ω$ and gives full scale deflection for a current of 4 $m A$ . To convert it to an ammeter of range 0 to $6 A$

1

10 m Ω

resistance is to be connected in parallel to the galvanometer

2

10 m Ω

resistance is to be connected in series with the galvanometer

3

0.1 Ω

resistance is to be connected in parallel to the galvanometer

4

0.1 Ω

resistance is to be connected in series with the galvanometer

Explanation:

A $G = 15 Ω$
$I_{g} = 4 mA = 4 \times 10^{- 3} Amp$
Total current $(I) = 6 A$
Formula used,
$S = \frac{I_{g} G}{I - I_{g}}$
$S = \frac{4 \times 10^{- 3} \times 15}{6 - 4 \times 10^{- 3}}$
$S = \frac{4 \times 10^{- 3} \times 15 \times 10^{3}}{6000 - 4}$
$S = \frac{60}{5996}$
$S = 0.0100$
$S = 10 \times 10^{- 3} Ω$
$S = 10 m Ω$
Therefore a resistance of $10 m Ω$ should be connected in parallel to the galvanometer.

JCECE-2009

Magnetism and Matter

154249 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 Resistance of an ammeter $(R_{A}) = 13 Ω$
Current flowing through $R_{A}$ is $I_{A} = 100 A$
Let Shunt resistance be $R_{S}$
$∴$ Current flowing through $R_{S}$ ,
$(I_{S}) = 750 - I_{A}$
$I_{S} = 750 - 100$
$I_{S} = 650 A$
Potential difference across ammeter and shunt is same
$I_{S} \cdot R_{S} = I_{A} \cdot R_{A}$
$650 \times R_{S} = 100 \times 13$
$R_{S} = 2 Ω$

JCECE-2008

Magnetism and Matter

154250 In an ammeter, $4 %$ of the main current is passing through galvanometer. If the galvanometer is shunted with a $5 Ω$ resistance, the resistance of the galvanometer is :

1

120 Ω

2

20 Ω

3

5 Ω

4

4 Ω

Explanation:

A Given,
$I_{G} = 4 % of I, I_{G} = \frac{4}{100} I$
$\frac{I_{G}}{I} = \frac{4}{100}$
$I_{G} \times R_{G} = I_{S} \times R_{S}$
$I_{G} \times R_{G} = (I - I_{G}) \times R_{S}$
$R_{G} = (\frac{I}{I_{G}} - 1) R_{S}$
$∴ R_{G} = (\frac{100}{4} - 1) 5 = (25 - 1) 5 = 120 Ω$

JCECE-2004

Magnetism and Matter

154246 The radius of the coil of a TG which has 10 turns is $0.1 m$ . The current required to produce a deflection of $60^{\circ} (B_{h} = 4 \times 10^{- 5} T)$ is:

1

3 A

2

1.1 A

3

2.1 A

4

1.5 A

5

2.6 A

Explanation:

B Given, radius of coil $= 0.1 m, θ = 60^{\circ}$
$B = B_{H} \tan θ$
$\frac{μ_{o} ni}{2 r} = B_{H} \tan θ$
$i = \frac{2 {rB}_{H} \tan θ}{μ_{o} n} = \frac{2 \times 0.1 \times 4 \times 10^{- 5} \times \tan 60^{\circ}}{4 π \times 10^{- 7} \times 10}$
$i = \frac{2 \times 0.1 \times 4 \times 10^{- 5} \times \sqrt{3}}{10 \times 4 π \times 10^{- 7}}$
$i = 1.1 Amp$

Kerala CEE 2004

Magnetism and Matter

154247 A $50 Ω$ galvanometer is shunted by a resistance of $5 Ω$ . The percentage of the total current, which passes through the galvanometer is

1

8.1 %

2

10.1 %

3

11.1 %

4

9.1 %

Explanation:

D Given, $R_{G} = 50 Ω, R_{S} = 5 Ω$
$I_{G} \times R_{G} = (I - I_{G}) R_{S}$
$I_{G} \times 50 = (I - I_{G}) 5$
$10 I_{G} = I - I_{G}$
$I = 11 I_{G}$
$I_{G} = \frac{I}{11}$
Let $%$ of current passing through galvanometer is $x$ $x %$ of $I = I_{G}$
$\frac{x}{100} \times I = \frac{I}{11}$
$x = \frac{100}{11} = 9.09 = 9.1 %$

JCECE-2015

Magnetism and Matter

154248 A galvanometer coil has a resistance of $15 Ω$ and gives full scale deflection for a current of 4 $m A$ . To convert it to an ammeter of range 0 to $6 A$

1

10 m Ω

resistance is to be connected in parallel to the galvanometer

2

10 m Ω

resistance is to be connected in series with the galvanometer

3

0.1 Ω

resistance is to be connected in parallel to the galvanometer

4

0.1 Ω

resistance is to be connected in series with the galvanometer

Explanation:

A $G = 15 Ω$
$I_{g} = 4 mA = 4 \times 10^{- 3} Amp$
Total current $(I) = 6 A$
Formula used,
$S = \frac{I_{g} G}{I - I_{g}}$
$S = \frac{4 \times 10^{- 3} \times 15}{6 - 4 \times 10^{- 3}}$
$S = \frac{4 \times 10^{- 3} \times 15 \times 10^{3}}{6000 - 4}$
$S = \frac{60}{5996}$
$S = 0.0100$
$S = 10 \times 10^{- 3} Ω$
$S = 10 m Ω$
Therefore a resistance of $10 m Ω$ should be connected in parallel to the galvanometer.

JCECE-2009

Magnetism and Matter

154249 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 Resistance of an ammeter $(R_{A}) = 13 Ω$
Current flowing through $R_{A}$ is $I_{A} = 100 A$
Let Shunt resistance be $R_{S}$
$∴$ Current flowing through $R_{S}$ ,
$(I_{S}) = 750 - I_{A}$
$I_{S} = 750 - 100$
$I_{S} = 650 A$
Potential difference across ammeter and shunt is same
$I_{S} \cdot R_{S} = I_{A} \cdot R_{A}$
$650 \times R_{S} = 100 \times 13$
$R_{S} = 2 Ω$

JCECE-2008

Magnetism and Matter

154250 In an ammeter, $4 %$ of the main current is passing through galvanometer. If the galvanometer is shunted with a $5 Ω$ resistance, the resistance of the galvanometer is :

1

120 Ω

2

20 Ω

3

5 Ω

4

4 Ω

Explanation:

A Given,
$I_{G} = 4 % of I, I_{G} = \frac{4}{100} I$
$\frac{I_{G}}{I} = \frac{4}{100}$
$I_{G} \times R_{G} = I_{S} \times R_{S}$
$I_{G} \times R_{G} = (I - I_{G}) \times R_{S}$
$R_{G} = (\frac{I}{I_{G}} - 1) R_{S}$
$∴ R_{G} = (\frac{100}{4} - 1) 5 = (25 - 1) 5 = 120 Ω$

JCECE-2004

Magnetism and Matter

154246 The radius of the coil of a TG which has 10 turns is $0.1 m$ . The current required to produce a deflection of $60^{\circ} (B_{h} = 4 \times 10^{- 5} T)$ is:

1

3 A

2

1.1 A

3

2.1 A

4

1.5 A

5

2.6 A

Explanation:

B Given, radius of coil $= 0.1 m, θ = 60^{\circ}$
$B = B_{H} \tan θ$
$\frac{μ_{o} ni}{2 r} = B_{H} \tan θ$
$i = \frac{2 {rB}_{H} \tan θ}{μ_{o} n} = \frac{2 \times 0.1 \times 4 \times 10^{- 5} \times \tan 60^{\circ}}{4 π \times 10^{- 7} \times 10}$
$i = \frac{2 \times 0.1 \times 4 \times 10^{- 5} \times \sqrt{3}}{10 \times 4 π \times 10^{- 7}}$
$i = 1.1 Amp$

Kerala CEE 2004

Magnetism and Matter

154247 A $50 Ω$ galvanometer is shunted by a resistance of $5 Ω$ . The percentage of the total current, which passes through the galvanometer is

1

8.1 %

2

10.1 %

3

11.1 %

4

9.1 %

Explanation:

D Given, $R_{G} = 50 Ω, R_{S} = 5 Ω$
$I_{G} \times R_{G} = (I - I_{G}) R_{S}$
$I_{G} \times 50 = (I - I_{G}) 5$
$10 I_{G} = I - I_{G}$
$I = 11 I_{G}$
$I_{G} = \frac{I}{11}$
Let $%$ of current passing through galvanometer is $x$ $x %$ of $I = I_{G}$
$\frac{x}{100} \times I = \frac{I}{11}$
$x = \frac{100}{11} = 9.09 = 9.1 %$

JCECE-2015

Magnetism and Matter

154248 A galvanometer coil has a resistance of $15 Ω$ and gives full scale deflection for a current of 4 $m A$ . To convert it to an ammeter of range 0 to $6 A$

1

10 m Ω

resistance is to be connected in parallel to the galvanometer

2

10 m Ω

resistance is to be connected in series with the galvanometer

3

0.1 Ω

resistance is to be connected in parallel to the galvanometer

4

0.1 Ω

resistance is to be connected in series with the galvanometer

Explanation:

A $G = 15 Ω$
$I_{g} = 4 mA = 4 \times 10^{- 3} Amp$
Total current $(I) = 6 A$
Formula used,
$S = \frac{I_{g} G}{I - I_{g}}$
$S = \frac{4 \times 10^{- 3} \times 15}{6 - 4 \times 10^{- 3}}$
$S = \frac{4 \times 10^{- 3} \times 15 \times 10^{3}}{6000 - 4}$
$S = \frac{60}{5996}$
$S = 0.0100$
$S = 10 \times 10^{- 3} Ω$
$S = 10 m Ω$
Therefore a resistance of $10 m Ω$ should be connected in parallel to the galvanometer.

JCECE-2009

Magnetism and Matter

154249 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 Resistance of an ammeter $(R_{A}) = 13 Ω$
Current flowing through $R_{A}$ is $I_{A} = 100 A$
Let Shunt resistance be $R_{S}$
$∴$ Current flowing through $R_{S}$ ,
$(I_{S}) = 750 - I_{A}$
$I_{S} = 750 - 100$
$I_{S} = 650 A$
Potential difference across ammeter and shunt is same
$I_{S} \cdot R_{S} = I_{A} \cdot R_{A}$
$650 \times R_{S} = 100 \times 13$
$R_{S} = 2 Ω$

JCECE-2008

Magnetism and Matter

154250 In an ammeter, $4 %$ of the main current is passing through galvanometer. If the galvanometer is shunted with a $5 Ω$ resistance, the resistance of the galvanometer is :

1

120 Ω

2

20 Ω

3

5 Ω

4

4 Ω

Explanation:

A Given,
$I_{G} = 4 % of I, I_{G} = \frac{4}{100} I$
$\frac{I_{G}}{I} = \frac{4}{100}$
$I_{G} \times R_{G} = I_{S} \times R_{S}$
$I_{G} \times R_{G} = (I - I_{G}) \times R_{S}$
$R_{G} = (\frac{I}{I_{G}} - 1) R_{S}$
$∴ R_{G} = (\frac{100}{4} - 1) 5 = (25 - 1) 5 = 120 Ω$

JCECE-2004