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Magnetism and Matter

154242 If the ammeter $A$ shows a zero reading in the circuit shown below, the value of resistance $R$ is

1

500 Ω

2

125 Ω

3

100 Ω

4

41.5 Ω

5

4 Ω

Explanation:

B
Apply KVL in loop 1,
$10 - 500 I - RI = 0$
$10 = I (500 + R)$
Apply KVL in loop 2,
$10 - 500 I - 2 = 0$
$500 I = 8$
$I = \frac{8}{500}$
Putting the value of I in equation (i),
$10 = \frac{8}{500} (500 + R)$
$10 = 8 + \frac{8}{500} R$
$2 = \frac{8}{500} R$
$R = \frac{2 \times 500}{8}$
$R = 125$

Kerala CEE- 2014

Magnetism and Matter

154243 The shunt resistance required to allow $4 %$ of the main current through the galvanometer of resistance $48 Ω$ is

1

1 Ω

2

2 Ω

3

3 Ω

4

4 Ω

5

5 Ω

Explanation:

B Current through the galvanometer is $4 %$ of main current $I_{G} = \frac{4 I}{100} = \frac{I}{25}$
Resistance of galvanometer $(G) = 48 Ω$
$R_{S} = \frac{{GI}_{G}}{I - I_{G}}$
$R_{S} = \frac{48 \times \frac{I}{25}}{I - \frac{I}{25}}$
$R_{S} = \frac{48}{24} = 2$
$R_{S} = 2 Ω$

Kerala CEE 2012

Magnetism and Matter

154244 A galvanometer of resistance $100 Ω$ is converted to a voltmeter of range $10 V$ by connecting a resistance of $10 k Ω$ . The resistance required to convert the same galvanometer to an ammeter of range $1 A$ is

1

0.4 Ω

2

0.3 Ω

3

1.2 Ω

4

0.2 Ω

5

0.1 Ω

Explanation:

E Given that, $G = 100 Ω$
$V = 10 V$
$R = 10 k Ω$
$I_{G} = \frac{V}{G + R}$
$I_{G} = \frac{10}{100 + 10 \times 10^{3}}$
$I_{G} = \frac{1}{1010}$
When shunt $S$ is connected in parallel
$S = \frac{{GI}_{G}}{I - I_{G}}$
$S = \frac{100 \times \frac{1}{1010}}{1 - \frac{1}{1010}}$
$S = \frac{100}{1009}$
$S = 0.099 \approx 0.1 Ω$

Kerala CEE - 2010

Magnetism and Matter

154245 A galvanometer of resistance $20 Ω$ shows a deflection of 10 division when a current of 1 $mA$ is passed through it. If a shunt of $4 Ω$ is connected and there are 50 division on the scale, the range of the galvanometer is

1

1 A

2

3 A

3

10 mA

4

30 A

5

30 mA

Explanation:

E Current for 10 division $= 1 mA$
Current for 50 division
$I_{g} = \frac{1 \times 50}{10} = 5 mA$
$I_{g} G = (I - I_{g}) S$
$5 \times 10^{- 3} \times 20 = (I - 5 \times 10^{- 3}) 4$
$25 \times 10^{- 3} = I - 5 \times 10^{- 3}$
$I = 30 \times 10^{- 3}$
$I = 30 mA$

Kerala CEE 2007

Magnetism and Matter

154242 If the ammeter $A$ shows a zero reading in the circuit shown below, the value of resistance $R$ is

1

500 Ω

2

125 Ω

3

100 Ω

4

41.5 Ω

5

4 Ω

Explanation:

B
Apply KVL in loop 1,
$10 - 500 I - RI = 0$
$10 = I (500 + R)$
Apply KVL in loop 2,
$10 - 500 I - 2 = 0$
$500 I = 8$
$I = \frac{8}{500}$
Putting the value of I in equation (i),
$10 = \frac{8}{500} (500 + R)$
$10 = 8 + \frac{8}{500} R$
$2 = \frac{8}{500} R$
$R = \frac{2 \times 500}{8}$
$R = 125$

Kerala CEE- 2014

Magnetism and Matter

154243 The shunt resistance required to allow $4 %$ of the main current through the galvanometer of resistance $48 Ω$ is

1

1 Ω

2

2 Ω

3

3 Ω

4

4 Ω

5

5 Ω

Explanation:

B Current through the galvanometer is $4 %$ of main current $I_{G} = \frac{4 I}{100} = \frac{I}{25}$
Resistance of galvanometer $(G) = 48 Ω$
$R_{S} = \frac{{GI}_{G}}{I - I_{G}}$
$R_{S} = \frac{48 \times \frac{I}{25}}{I - \frac{I}{25}}$
$R_{S} = \frac{48}{24} = 2$
$R_{S} = 2 Ω$

Kerala CEE 2012

Magnetism and Matter

154244 A galvanometer of resistance $100 Ω$ is converted to a voltmeter of range $10 V$ by connecting a resistance of $10 k Ω$ . The resistance required to convert the same galvanometer to an ammeter of range $1 A$ is

1

0.4 Ω

2

0.3 Ω

3

1.2 Ω

4

0.2 Ω

5

0.1 Ω

Explanation:

E Given that, $G = 100 Ω$
$V = 10 V$
$R = 10 k Ω$
$I_{G} = \frac{V}{G + R}$
$I_{G} = \frac{10}{100 + 10 \times 10^{3}}$
$I_{G} = \frac{1}{1010}$
When shunt $S$ is connected in parallel
$S = \frac{{GI}_{G}}{I - I_{G}}$
$S = \frac{100 \times \frac{1}{1010}}{1 - \frac{1}{1010}}$
$S = \frac{100}{1009}$
$S = 0.099 \approx 0.1 Ω$

Kerala CEE - 2010

Magnetism and Matter

154245 A galvanometer of resistance $20 Ω$ shows a deflection of 10 division when a current of 1 $mA$ is passed through it. If a shunt of $4 Ω$ is connected and there are 50 division on the scale, the range of the galvanometer is

1

1 A

2

3 A

3

10 mA

4

30 A

5

30 mA

Explanation:

E Current for 10 division $= 1 mA$
Current for 50 division
$I_{g} = \frac{1 \times 50}{10} = 5 mA$
$I_{g} G = (I - I_{g}) S$
$5 \times 10^{- 3} \times 20 = (I - 5 \times 10^{- 3}) 4$
$25 \times 10^{- 3} = I - 5 \times 10^{- 3}$
$I = 30 \times 10^{- 3}$
$I = 30 mA$

Kerala CEE 2007

Magnetism and Matter

154242 If the ammeter $A$ shows a zero reading in the circuit shown below, the value of resistance $R$ is

1

500 Ω

2

125 Ω

3

100 Ω

4

41.5 Ω

5

4 Ω

Explanation:

B
Apply KVL in loop 1,
$10 - 500 I - RI = 0$
$10 = I (500 + R)$
Apply KVL in loop 2,
$10 - 500 I - 2 = 0$
$500 I = 8$
$I = \frac{8}{500}$
Putting the value of I in equation (i),
$10 = \frac{8}{500} (500 + R)$
$10 = 8 + \frac{8}{500} R$
$2 = \frac{8}{500} R$
$R = \frac{2 \times 500}{8}$
$R = 125$

Kerala CEE- 2014

Magnetism and Matter

154243 The shunt resistance required to allow $4 %$ of the main current through the galvanometer of resistance $48 Ω$ is

1

1 Ω

2

2 Ω

3

3 Ω

4

4 Ω

5

5 Ω

Explanation:

B Current through the galvanometer is $4 %$ of main current $I_{G} = \frac{4 I}{100} = \frac{I}{25}$
Resistance of galvanometer $(G) = 48 Ω$
$R_{S} = \frac{{GI}_{G}}{I - I_{G}}$
$R_{S} = \frac{48 \times \frac{I}{25}}{I - \frac{I}{25}}$
$R_{S} = \frac{48}{24} = 2$
$R_{S} = 2 Ω$

Kerala CEE 2012

Magnetism and Matter

154244 A galvanometer of resistance $100 Ω$ is converted to a voltmeter of range $10 V$ by connecting a resistance of $10 k Ω$ . The resistance required to convert the same galvanometer to an ammeter of range $1 A$ is

1

0.4 Ω

2

0.3 Ω

3

1.2 Ω

4

0.2 Ω

5

0.1 Ω

Explanation:

E Given that, $G = 100 Ω$
$V = 10 V$
$R = 10 k Ω$
$I_{G} = \frac{V}{G + R}$
$I_{G} = \frac{10}{100 + 10 \times 10^{3}}$
$I_{G} = \frac{1}{1010}$
When shunt $S$ is connected in parallel
$S = \frac{{GI}_{G}}{I - I_{G}}$
$S = \frac{100 \times \frac{1}{1010}}{1 - \frac{1}{1010}}$
$S = \frac{100}{1009}$
$S = 0.099 \approx 0.1 Ω$

Kerala CEE - 2010

Magnetism and Matter

154245 A galvanometer of resistance $20 Ω$ shows a deflection of 10 division when a current of 1 $mA$ is passed through it. If a shunt of $4 Ω$ is connected and there are 50 division on the scale, the range of the galvanometer is

1

1 A

2

3 A

3

10 mA

4

30 A

5

30 mA

Explanation:

E Current for 10 division $= 1 mA$
Current for 50 division
$I_{g} = \frac{1 \times 50}{10} = 5 mA$
$I_{g} G = (I - I_{g}) S$
$5 \times 10^{- 3} \times 20 = (I - 5 \times 10^{- 3}) 4$
$25 \times 10^{- 3} = I - 5 \times 10^{- 3}$
$I = 30 \times 10^{- 3}$
$I = 30 mA$

Kerala CEE 2007

Magnetism and Matter

154242 If the ammeter $A$ shows a zero reading in the circuit shown below, the value of resistance $R$ is

1

500 Ω

2

125 Ω

3

100 Ω

4

41.5 Ω

5

4 Ω

Explanation:

B
Apply KVL in loop 1,
$10 - 500 I - RI = 0$
$10 = I (500 + R)$
Apply KVL in loop 2,
$10 - 500 I - 2 = 0$
$500 I = 8$
$I = \frac{8}{500}$
Putting the value of I in equation (i),
$10 = \frac{8}{500} (500 + R)$
$10 = 8 + \frac{8}{500} R$
$2 = \frac{8}{500} R$
$R = \frac{2 \times 500}{8}$
$R = 125$

Kerala CEE- 2014

Magnetism and Matter

154243 The shunt resistance required to allow $4 %$ of the main current through the galvanometer of resistance $48 Ω$ is

1

1 Ω

2

2 Ω

3

3 Ω

4

4 Ω

5

5 Ω

Explanation:

B Current through the galvanometer is $4 %$ of main current $I_{G} = \frac{4 I}{100} = \frac{I}{25}$
Resistance of galvanometer $(G) = 48 Ω$
$R_{S} = \frac{{GI}_{G}}{I - I_{G}}$
$R_{S} = \frac{48 \times \frac{I}{25}}{I - \frac{I}{25}}$
$R_{S} = \frac{48}{24} = 2$
$R_{S} = 2 Ω$

Kerala CEE 2012

Magnetism and Matter

154244 A galvanometer of resistance $100 Ω$ is converted to a voltmeter of range $10 V$ by connecting a resistance of $10 k Ω$ . The resistance required to convert the same galvanometer to an ammeter of range $1 A$ is

1

0.4 Ω

2

0.3 Ω

3

1.2 Ω

4

0.2 Ω

5

0.1 Ω

Explanation:

E Given that, $G = 100 Ω$
$V = 10 V$
$R = 10 k Ω$
$I_{G} = \frac{V}{G + R}$
$I_{G} = \frac{10}{100 + 10 \times 10^{3}}$
$I_{G} = \frac{1}{1010}$
When shunt $S$ is connected in parallel
$S = \frac{{GI}_{G}}{I - I_{G}}$
$S = \frac{100 \times \frac{1}{1010}}{1 - \frac{1}{1010}}$
$S = \frac{100}{1009}$
$S = 0.099 \approx 0.1 Ω$

Kerala CEE - 2010

Magnetism and Matter

154245 A galvanometer of resistance $20 Ω$ shows a deflection of 10 division when a current of 1 $mA$ is passed through it. If a shunt of $4 Ω$ is connected and there are 50 division on the scale, the range of the galvanometer is

1

1 A

2

3 A

3

10 mA

4

30 A

5

30 mA

Explanation:

E Current for 10 division $= 1 mA$
Current for 50 division
$I_{g} = \frac{1 \times 50}{10} = 5 mA$
$I_{g} G = (I - I_{g}) S$
$5 \times 10^{- 3} \times 20 = (I - 5 \times 10^{- 3}) 4$
$25 \times 10^{- 3} = I - 5 \times 10^{- 3}$
$I = 30 \times 10^{- 3}$
$I = 30 mA$

Kerala CEE 2007

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