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PHXII04:MOVING CHARGES AND MAGNETISM

362968 A galvanometer may be converted into ammeter or a voltmeter.The resistance of the device so obtained will be the largest in case of

1 Ammeter of range 1

A

2 Ammeter of range 10

A

3 Voltmeter of range 1

V

4 Voltmeter of range 10

V

PHXII04:MOVING CHARGES AND MAGNETISM

362969 Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across $A$ and $B$ , their readings are $V_{1}, V_{2}$ and $V_{3}$ . Choose the correct options.
supporting img

1

V_{1} \neq V_{3} - V_{2}

2

V_{1} + V_{2} = V_{3}

3

V_{1} = V_{2}

4

V_{1} + V_{2} > V_{2}

Explanation:

Using ohm's law across points $A$ and $B$
$V = i R$
$V_{A B} = I_{1} (R_{1} + R_{2})$
$V_{A B} = I_{2} R_{3}$
supporting img
$I_{1} (R_{1} + R_{2}) = I_{2} R_{3}$
$\Rightarrow V_{1} + V_{2} = V_{3}$

JEE - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362970 A voltmeter having resistance of $50 \times 10^{3} o h m$ is used to measure the voltage in a circuit. To increase the range of measurement 3 times the additional series resistance required is

1

10^{5} o h m

2

9 \times 10^{6} o h m

3

150 k . o h m

4

900 k . o h m

Explanation:

Given that $R_{v} = 50 \times 10^{3} o h m$
The range of a voltmeter is
$V = i_{g} R_{v}$
As the range is to be made three times
$3 V = i_{g} (R_{v} + S)$
From both the equations we get
$S = 10^{5} Ω$

PHXII04:MOVING CHARGES AND MAGNETISM

362971 A galvanometer has a resistance of $50 Ω$ and it allows maximum current of $5 m A .$ It can be converted into voltmeter to measure upto $100 V$ by connecting it in series to a resistor of resistance

1

20050 Ω

2

19500 Ω

3

5975 Ω

4

19950 Ω

Explanation:

supporting img

Given, $R_{G} = 50 Ω, I = 5 m A, V = 100 V$
Let the resistance is $R$ .
$(R + R_{G}) I = V$
$(R + 50) \times 5 \times 10^{- 3} = 100$
$R = 20000 - 50 = 19950 Ω$

JEE - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362968 A galvanometer may be converted into ammeter or a voltmeter.The resistance of the device so obtained will be the largest in case of

1 Ammeter of range 1

A

2 Ammeter of range 10

A

3 Voltmeter of range 1

V

4 Voltmeter of range 10

V

PHXII04:MOVING CHARGES AND MAGNETISM

362969 Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across $A$ and $B$ , their readings are $V_{1}, V_{2}$ and $V_{3}$ . Choose the correct options.
supporting img

1

V_{1} \neq V_{3} - V_{2}

2

V_{1} + V_{2} = V_{3}

3

V_{1} = V_{2}

4

V_{1} + V_{2} > V_{2}

Explanation:

Using ohm's law across points $A$ and $B$
$V = i R$
$V_{A B} = I_{1} (R_{1} + R_{2})$
$V_{A B} = I_{2} R_{3}$
supporting img
$I_{1} (R_{1} + R_{2}) = I_{2} R_{3}$
$\Rightarrow V_{1} + V_{2} = V_{3}$

JEE - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362970 A voltmeter having resistance of $50 \times 10^{3} o h m$ is used to measure the voltage in a circuit. To increase the range of measurement 3 times the additional series resistance required is

1

10^{5} o h m

2

9 \times 10^{6} o h m

3

150 k . o h m

4

900 k . o h m

Explanation:

Given that $R_{v} = 50 \times 10^{3} o h m$
The range of a voltmeter is
$V = i_{g} R_{v}$
As the range is to be made three times
$3 V = i_{g} (R_{v} + S)$
From both the equations we get
$S = 10^{5} Ω$

PHXII04:MOVING CHARGES AND MAGNETISM

362971 A galvanometer has a resistance of $50 Ω$ and it allows maximum current of $5 m A .$ It can be converted into voltmeter to measure upto $100 V$ by connecting it in series to a resistor of resistance

1

20050 Ω

2

19500 Ω

3

5975 Ω

4

19950 Ω

Explanation:

supporting img

Given, $R_{G} = 50 Ω, I = 5 m A, V = 100 V$
Let the resistance is $R$ .
$(R + R_{G}) I = V$
$(R + 50) \times 5 \times 10^{- 3} = 100$
$R = 20000 - 50 = 19950 Ω$

JEE - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362968 A galvanometer may be converted into ammeter or a voltmeter.The resistance of the device so obtained will be the largest in case of

1 Ammeter of range 1

A

2 Ammeter of range 10

A

3 Voltmeter of range 1

V

4 Voltmeter of range 10

V

PHXII04:MOVING CHARGES AND MAGNETISM

362969 Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across $A$ and $B$ , their readings are $V_{1}, V_{2}$ and $V_{3}$ . Choose the correct options.
supporting img

1

V_{1} \neq V_{3} - V_{2}

2

V_{1} + V_{2} = V_{3}

3

V_{1} = V_{2}

4

V_{1} + V_{2} > V_{2}

Explanation:

Using ohm's law across points $A$ and $B$
$V = i R$
$V_{A B} = I_{1} (R_{1} + R_{2})$
$V_{A B} = I_{2} R_{3}$
supporting img
$I_{1} (R_{1} + R_{2}) = I_{2} R_{3}$
$\Rightarrow V_{1} + V_{2} = V_{3}$

JEE - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362970 A voltmeter having resistance of $50 \times 10^{3} o h m$ is used to measure the voltage in a circuit. To increase the range of measurement 3 times the additional series resistance required is

1

10^{5} o h m

2

9 \times 10^{6} o h m

3

150 k . o h m

4

900 k . o h m

Explanation:

Given that $R_{v} = 50 \times 10^{3} o h m$
The range of a voltmeter is
$V = i_{g} R_{v}$
As the range is to be made three times
$3 V = i_{g} (R_{v} + S)$
From both the equations we get
$S = 10^{5} Ω$

PHXII04:MOVING CHARGES AND MAGNETISM

362971 A galvanometer has a resistance of $50 Ω$ and it allows maximum current of $5 m A .$ It can be converted into voltmeter to measure upto $100 V$ by connecting it in series to a resistor of resistance

1

20050 Ω

2

19500 Ω

3

5975 Ω

4

19950 Ω

Explanation:

supporting img

Given, $R_{G} = 50 Ω, I = 5 m A, V = 100 V$
Let the resistance is $R$ .
$(R + R_{G}) I = V$
$(R + 50) \times 5 \times 10^{- 3} = 100$
$R = 20000 - 50 = 19950 Ω$

JEE - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362968 A galvanometer may be converted into ammeter or a voltmeter.The resistance of the device so obtained will be the largest in case of

1 Ammeter of range 1

A

2 Ammeter of range 10

A

3 Voltmeter of range 1

V

4 Voltmeter of range 10

V

PHXII04:MOVING CHARGES AND MAGNETISM

362969 Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across $A$ and $B$ , their readings are $V_{1}, V_{2}$ and $V_{3}$ . Choose the correct options.
supporting img

1

V_{1} \neq V_{3} - V_{2}

2

V_{1} + V_{2} = V_{3}

3

V_{1} = V_{2}

4

V_{1} + V_{2} > V_{2}

Explanation:

Using ohm's law across points $A$ and $B$
$V = i R$
$V_{A B} = I_{1} (R_{1} + R_{2})$
$V_{A B} = I_{2} R_{3}$
supporting img
$I_{1} (R_{1} + R_{2}) = I_{2} R_{3}$
$\Rightarrow V_{1} + V_{2} = V_{3}$

JEE - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362970 A voltmeter having resistance of $50 \times 10^{3} o h m$ is used to measure the voltage in a circuit. To increase the range of measurement 3 times the additional series resistance required is

1

10^{5} o h m

2

9 \times 10^{6} o h m

3

150 k . o h m

4

900 k . o h m

Explanation:

Given that $R_{v} = 50 \times 10^{3} o h m$
The range of a voltmeter is
$V = i_{g} R_{v}$
As the range is to be made three times
$3 V = i_{g} (R_{v} + S)$
From both the equations we get
$S = 10^{5} Ω$

PHXII04:MOVING CHARGES AND MAGNETISM

362971 A galvanometer has a resistance of $50 Ω$ and it allows maximum current of $5 m A .$ It can be converted into voltmeter to measure upto $100 V$ by connecting it in series to a resistor of resistance

1

20050 Ω

2

19500 Ω

3

5975 Ω

4

19950 Ω

Explanation:

supporting img

Given, $R_{G} = 50 Ω, I = 5 m A, V = 100 V$
Let the resistance is $R$ .
$(R + R_{G}) I = V$
$(R + 50) \times 5 \times 10^{- 3} = 100$
$R = 20000 - 50 = 19950 Ω$

JEE - 2024

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