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PHXII03:CURRENT ELECTRICITY

357256 A voltmeter of a very high resistance is joined in the circuit as shown in figure. The voltage shown by this voltmeter will be
supporting img

1

6 V

2

3 V

3

5 V

4

2.5 V

Explanation:

The net current through the circuit is
$i = \frac{10}{6 + \frac{8 \times 8}{8 + 8}} = 1 A$
The reading of voltmeter $= (6 Ω) (1 A) = 6 V$

PHXII03:CURRENT ELECTRICITY

357257 In the figure shown, the reading of the ideal voltmeter is 0 . The relation between $R, r_{1}$ and $r_{2}$ is
supporting img

1

R = r_{2} - r_{1}

2

R = r_{1} - r_{2}

3

R = r_{1} + r_{2}

4

R = \frac{r_{1} r_{2}}{r_{1} + r_{2}}

Explanation:

Here, $V = E - I r_{1} \Rightarrow 0 = E - I r_{1} (∵ V = 0)$
$∴ I = \frac{E}{r_{1}} (1)$
Current in circuit, $I = \frac{E + E}{r_{1} + r_{2} + R}$
$I = \frac{2 E}{r_{1} + r_{2} + R} (2)$
From eqns. (1) and (2) we get $\begin{aligned} \frac{E}{r_{1}} = \frac{2 E}{r_{1} + r_{2} + R} \Rightarrow 2 r_{1} = r_{1} + r_{2} + R \\ ∴ R = r_{1} - r_{2} \end{aligned}$ .
So correct option is (2)

PHXII03:CURRENT ELECTRICITY

357258 A 10 $V$ battery with internal resistance $1 Ω$ and 15 $V$ battery with internal resistance $0.6 Ω$ are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to :
supporting img

1 11.9

V

2 12.5

V

3 24.5

V

4 13.1

V

Explanation:

The current through the upper loop is
supporting img
$i = \frac{5}{1.6} = \frac{25}{8} A m p$
The voltmeter reading is
$V = 15 - \frac{25}{8} \times 0.6 = 15 - \frac{15}{8} = 13.1 V$

JEE - 2015

PHXII03:CURRENT ELECTRICITY

357259 Assertion :
Voltameter measures current more accurately than ammeter.
Reason :
Relative error will be small if measured from voltameter.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Voltameter measures current in directly in terms of mass $m$ of ions deposited and electro chemical equivalant $Z$ of substance. Then, values of $m, Z$ can be given to 3 rd decimal place and 5th place. $\Rightarrow$ Assertion is true and reason describes it. So correct option is (1).

PHXII03:CURRENT ELECTRICITY

357256 A voltmeter of a very high resistance is joined in the circuit as shown in figure. The voltage shown by this voltmeter will be
supporting img

1

6 V

2

3 V

3

5 V

4

2.5 V

Explanation:

The net current through the circuit is
$i = \frac{10}{6 + \frac{8 \times 8}{8 + 8}} = 1 A$
The reading of voltmeter $= (6 Ω) (1 A) = 6 V$

PHXII03:CURRENT ELECTRICITY

357257 In the figure shown, the reading of the ideal voltmeter is 0 . The relation between $R, r_{1}$ and $r_{2}$ is
supporting img

1

R = r_{2} - r_{1}

2

R = r_{1} - r_{2}

3

R = r_{1} + r_{2}

4

R = \frac{r_{1} r_{2}}{r_{1} + r_{2}}

Explanation:

Here, $V = E - I r_{1} \Rightarrow 0 = E - I r_{1} (∵ V = 0)$
$∴ I = \frac{E}{r_{1}} (1)$
Current in circuit, $I = \frac{E + E}{r_{1} + r_{2} + R}$
$I = \frac{2 E}{r_{1} + r_{2} + R} (2)$
From eqns. (1) and (2) we get $\begin{aligned} \frac{E}{r_{1}} = \frac{2 E}{r_{1} + r_{2} + R} \Rightarrow 2 r_{1} = r_{1} + r_{2} + R \\ ∴ R = r_{1} - r_{2} \end{aligned}$ .
So correct option is (2)

PHXII03:CURRENT ELECTRICITY

357258 A 10 $V$ battery with internal resistance $1 Ω$ and 15 $V$ battery with internal resistance $0.6 Ω$ are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to :
supporting img

1 11.9

V

2 12.5

V

3 24.5

V

4 13.1

V

Explanation:

The current through the upper loop is
supporting img
$i = \frac{5}{1.6} = \frac{25}{8} A m p$
The voltmeter reading is
$V = 15 - \frac{25}{8} \times 0.6 = 15 - \frac{15}{8} = 13.1 V$

JEE - 2015

PHXII03:CURRENT ELECTRICITY

357259 Assertion :
Voltameter measures current more accurately than ammeter.
Reason :
Relative error will be small if measured from voltameter.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Voltameter measures current in directly in terms of mass $m$ of ions deposited and electro chemical equivalant $Z$ of substance. Then, values of $m, Z$ can be given to 3 rd decimal place and 5th place. $\Rightarrow$ Assertion is true and reason describes it. So correct option is (1).

PHXII03:CURRENT ELECTRICITY

357256 A voltmeter of a very high resistance is joined in the circuit as shown in figure. The voltage shown by this voltmeter will be
supporting img

1

6 V

2

3 V

3

5 V

4

2.5 V

Explanation:

The net current through the circuit is
$i = \frac{10}{6 + \frac{8 \times 8}{8 + 8}} = 1 A$
The reading of voltmeter $= (6 Ω) (1 A) = 6 V$

PHXII03:CURRENT ELECTRICITY

357257 In the figure shown, the reading of the ideal voltmeter is 0 . The relation between $R, r_{1}$ and $r_{2}$ is
supporting img

1

R = r_{2} - r_{1}

2

R = r_{1} - r_{2}

3

R = r_{1} + r_{2}

4

R = \frac{r_{1} r_{2}}{r_{1} + r_{2}}

Explanation:

Here, $V = E - I r_{1} \Rightarrow 0 = E - I r_{1} (∵ V = 0)$
$∴ I = \frac{E}{r_{1}} (1)$
Current in circuit, $I = \frac{E + E}{r_{1} + r_{2} + R}$
$I = \frac{2 E}{r_{1} + r_{2} + R} (2)$
From eqns. (1) and (2) we get $\begin{aligned} \frac{E}{r_{1}} = \frac{2 E}{r_{1} + r_{2} + R} \Rightarrow 2 r_{1} = r_{1} + r_{2} + R \\ ∴ R = r_{1} - r_{2} \end{aligned}$ .
So correct option is (2)

PHXII03:CURRENT ELECTRICITY

357258 A 10 $V$ battery with internal resistance $1 Ω$ and 15 $V$ battery with internal resistance $0.6 Ω$ are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to :
supporting img

1 11.9

V

2 12.5

V

3 24.5

V

4 13.1

V

Explanation:

The current through the upper loop is
supporting img
$i = \frac{5}{1.6} = \frac{25}{8} A m p$
The voltmeter reading is
$V = 15 - \frac{25}{8} \times 0.6 = 15 - \frac{15}{8} = 13.1 V$

JEE - 2015

PHXII03:CURRENT ELECTRICITY

357259 Assertion :
Voltameter measures current more accurately than ammeter.
Reason :
Relative error will be small if measured from voltameter.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Voltameter measures current in directly in terms of mass $m$ of ions deposited and electro chemical equivalant $Z$ of substance. Then, values of $m, Z$ can be given to 3 rd decimal place and 5th place. $\Rightarrow$ Assertion is true and reason describes it. So correct option is (1).

PHXII03:CURRENT ELECTRICITY

357256 A voltmeter of a very high resistance is joined in the circuit as shown in figure. The voltage shown by this voltmeter will be
supporting img

1

6 V

2

3 V

3

5 V

4

2.5 V

Explanation:

The net current through the circuit is
$i = \frac{10}{6 + \frac{8 \times 8}{8 + 8}} = 1 A$
The reading of voltmeter $= (6 Ω) (1 A) = 6 V$

PHXII03:CURRENT ELECTRICITY

357257 In the figure shown, the reading of the ideal voltmeter is 0 . The relation between $R, r_{1}$ and $r_{2}$ is
supporting img

1

R = r_{2} - r_{1}

2

R = r_{1} - r_{2}

3

R = r_{1} + r_{2}

4

R = \frac{r_{1} r_{2}}{r_{1} + r_{2}}

Explanation:

Here, $V = E - I r_{1} \Rightarrow 0 = E - I r_{1} (∵ V = 0)$
$∴ I = \frac{E}{r_{1}} (1)$
Current in circuit, $I = \frac{E + E}{r_{1} + r_{2} + R}$
$I = \frac{2 E}{r_{1} + r_{2} + R} (2)$
From eqns. (1) and (2) we get $\begin{aligned} \frac{E}{r_{1}} = \frac{2 E}{r_{1} + r_{2} + R} \Rightarrow 2 r_{1} = r_{1} + r_{2} + R \\ ∴ R = r_{1} - r_{2} \end{aligned}$ .
So correct option is (2)

PHXII03:CURRENT ELECTRICITY

357258 A 10 $V$ battery with internal resistance $1 Ω$ and 15 $V$ battery with internal resistance $0.6 Ω$ are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to :
supporting img

1 11.9

V

2 12.5

V

3 24.5

V

4 13.1

V

Explanation:

The current through the upper loop is
supporting img
$i = \frac{5}{1.6} = \frac{25}{8} A m p$
The voltmeter reading is
$V = 15 - \frac{25}{8} \times 0.6 = 15 - \frac{15}{8} = 13.1 V$

JEE - 2015

PHXII03:CURRENT ELECTRICITY

357259 Assertion :
Voltameter measures current more accurately than ammeter.
Reason :
Relative error will be small if measured from voltameter.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Voltameter measures current in directly in terms of mass $m$ of ions deposited and electro chemical equivalant $Z$ of substance. Then, values of $m, Z$ can be given to 3 rd decimal place and 5th place. $\Rightarrow$ Assertion is true and reason describes it. So correct option is (1).

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