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PHXII03:CURRENT ELECTRICITY

357238 The reading of ideal voltmeter in the following circuit is
supporting img

1

3.4 V

2

5.0 V

3

2.4 V

4

4.0 V

Explanation:

As the voltmeter is ideal then given circuit is an open circuit, so reading of voltmeter is equal to the e.m.f. of cell, i.e., 5.0 $V$ .

PHXII03:CURRENT ELECTRICITY

357239 An ideal voltmeter and an ideal ammeter connected as shown in the figure. The reading of the voltmeter is
supporting img

1

6 V

2

9 V

3

3 V

4 Zero

Explanation:

Since ideal ammeter has zero resistance and potential drop across it, $V = 0$

PHXII03:CURRENT ELECTRICITY

357240 Read the Assertion and Reason carefully to mark the correct options given below:
Assertion :
For measuring the potential difference across a resistance of $600 Ω$ , the voltmeter with resistance $1000 Ω$ will be preferred over voltmeter with resistance $4000 Ω$ .
Reason :
Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.

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

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

3 Assertion is incorrect but Reason is correct.

4 Assertion is correct but reason is incorrect.

Explanation:

Assertion $A$ is incorrect because a voltmeter should have very high resistance. So that very small current flows through it.

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357241 Two batteries, one of emf $18 V$ and internal resistance $2 Ω$ , and the other of emf $12 V$ and internal resistance $1 Ω$ , are connected as shown. The voltmeter $V$ will record a reading of:
supporting img

1

15 V

2

30 V

3

14 V

4

18 V

Explanation:

Is clear that the two cells oppose each other hence, the effective emf in closed circuit is $18 - 12 = 6 V$ and net resistance is $1 + 2 = 3 Ω$ (because in the closed circuit the internal resistances of two cells are in series).
The current in circuit will be in direction of arrow shown in figure.
supporting img
$I = \frac{effective emf}{total resistance} = \frac{6}{3} = 2 A$
The potential difference across $V$ will be same as the terminal voltage of either cell.
Since, current is drawn from the cell of 18 volt, hence,
$\begin{aligned} V_{1} = E_{1} - i r_{1} \\ = 18 - (2 \times 2) = 18 - 4 = 14 V \end{aligned}$
Similarly, current enters in the cell of $12 V$ hence
$V_{2} = E_{2} + i r_{2}$
$= 12 + 2 \times 1 = 12 + 2 = 14 V$
Hence, $V = 14 V$

PHXII03:CURRENT ELECTRICITY

357242 A voltmeter with resistance $2500 Ω$ indicates a voltage of 125 $V$ in the circuit shown in figure. What is the series resistance to be connected to voltmeter so that it indicates 100 $V$ ? Consider ideal battery.
supporting img

1

625 Ω

2

120 Ω

3

550 Ω

4 Data are insufficient

Explanation:

As the battery is ideal emf should be 125 $V$ .
For second case: Let $R$ is the resistance to be connected in series with the voltmeter.
$(\frac{125}{2500 + R}) 2500 = 100 V$
$\Rightarrow R = 625 Ω$

PHXII03:CURRENT ELECTRICITY

357238 The reading of ideal voltmeter in the following circuit is
supporting img

1

3.4 V

2

5.0 V

3

2.4 V

4

4.0 V

Explanation:

As the voltmeter is ideal then given circuit is an open circuit, so reading of voltmeter is equal to the e.m.f. of cell, i.e., 5.0 $V$ .

PHXII03:CURRENT ELECTRICITY

357239 An ideal voltmeter and an ideal ammeter connected as shown in the figure. The reading of the voltmeter is
supporting img

1

6 V

2

9 V

3

3 V

4 Zero

Explanation:

Since ideal ammeter has zero resistance and potential drop across it, $V = 0$

PHXII03:CURRENT ELECTRICITY

357240 Read the Assertion and Reason carefully to mark the correct options given below:
Assertion :
For measuring the potential difference across a resistance of $600 Ω$ , the voltmeter with resistance $1000 Ω$ will be preferred over voltmeter with resistance $4000 Ω$ .
Reason :
Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.

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

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

3 Assertion is incorrect but Reason is correct.

4 Assertion is correct but reason is incorrect.

Explanation:

Assertion $A$ is incorrect because a voltmeter should have very high resistance. So that very small current flows through it.

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357241 Two batteries, one of emf $18 V$ and internal resistance $2 Ω$ , and the other of emf $12 V$ and internal resistance $1 Ω$ , are connected as shown. The voltmeter $V$ will record a reading of:
supporting img

1

15 V

2

30 V

3

14 V

4

18 V

Explanation:

Is clear that the two cells oppose each other hence, the effective emf in closed circuit is $18 - 12 = 6 V$ and net resistance is $1 + 2 = 3 Ω$ (because in the closed circuit the internal resistances of two cells are in series).
The current in circuit will be in direction of arrow shown in figure.
supporting img
$I = \frac{effective emf}{total resistance} = \frac{6}{3} = 2 A$
The potential difference across $V$ will be same as the terminal voltage of either cell.
Since, current is drawn from the cell of 18 volt, hence,
$\begin{aligned} V_{1} = E_{1} - i r_{1} \\ = 18 - (2 \times 2) = 18 - 4 = 14 V \end{aligned}$
Similarly, current enters in the cell of $12 V$ hence
$V_{2} = E_{2} + i r_{2}$
$= 12 + 2 \times 1 = 12 + 2 = 14 V$
Hence, $V = 14 V$

PHXII03:CURRENT ELECTRICITY

357242 A voltmeter with resistance $2500 Ω$ indicates a voltage of 125 $V$ in the circuit shown in figure. What is the series resistance to be connected to voltmeter so that it indicates 100 $V$ ? Consider ideal battery.
supporting img

1

625 Ω

2

120 Ω

3

550 Ω

4 Data are insufficient

Explanation:

As the battery is ideal emf should be 125 $V$ .
For second case: Let $R$ is the resistance to be connected in series with the voltmeter.
$(\frac{125}{2500 + R}) 2500 = 100 V$
$\Rightarrow R = 625 Ω$

PHXII03:CURRENT ELECTRICITY

357238 The reading of ideal voltmeter in the following circuit is
supporting img

1

3.4 V

2

5.0 V

3

2.4 V

4

4.0 V

Explanation:

As the voltmeter is ideal then given circuit is an open circuit, so reading of voltmeter is equal to the e.m.f. of cell, i.e., 5.0 $V$ .

PHXII03:CURRENT ELECTRICITY

357239 An ideal voltmeter and an ideal ammeter connected as shown in the figure. The reading of the voltmeter is
supporting img

1

6 V

2

9 V

3

3 V

4 Zero

Explanation:

Since ideal ammeter has zero resistance and potential drop across it, $V = 0$

PHXII03:CURRENT ELECTRICITY

357240 Read the Assertion and Reason carefully to mark the correct options given below:
Assertion :
For measuring the potential difference across a resistance of $600 Ω$ , the voltmeter with resistance $1000 Ω$ will be preferred over voltmeter with resistance $4000 Ω$ .
Reason :
Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.

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

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

3 Assertion is incorrect but Reason is correct.

4 Assertion is correct but reason is incorrect.

Explanation:

Assertion $A$ is incorrect because a voltmeter should have very high resistance. So that very small current flows through it.

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357241 Two batteries, one of emf $18 V$ and internal resistance $2 Ω$ , and the other of emf $12 V$ and internal resistance $1 Ω$ , are connected as shown. The voltmeter $V$ will record a reading of:
supporting img

1

15 V

2

30 V

3

14 V

4

18 V

Explanation:

Is clear that the two cells oppose each other hence, the effective emf in closed circuit is $18 - 12 = 6 V$ and net resistance is $1 + 2 = 3 Ω$ (because in the closed circuit the internal resistances of two cells are in series).
The current in circuit will be in direction of arrow shown in figure.
supporting img
$I = \frac{effective emf}{total resistance} = \frac{6}{3} = 2 A$
The potential difference across $V$ will be same as the terminal voltage of either cell.
Since, current is drawn from the cell of 18 volt, hence,
$\begin{aligned} V_{1} = E_{1} - i r_{1} \\ = 18 - (2 \times 2) = 18 - 4 = 14 V \end{aligned}$
Similarly, current enters in the cell of $12 V$ hence
$V_{2} = E_{2} + i r_{2}$
$= 12 + 2 \times 1 = 12 + 2 = 14 V$
Hence, $V = 14 V$

PHXII03:CURRENT ELECTRICITY

357242 A voltmeter with resistance $2500 Ω$ indicates a voltage of 125 $V$ in the circuit shown in figure. What is the series resistance to be connected to voltmeter so that it indicates 100 $V$ ? Consider ideal battery.
supporting img

1

625 Ω

2

120 Ω

3

550 Ω

4 Data are insufficient

Explanation:

As the battery is ideal emf should be 125 $V$ .
For second case: Let $R$ is the resistance to be connected in series with the voltmeter.
$(\frac{125}{2500 + R}) 2500 = 100 V$
$\Rightarrow R = 625 Ω$

PHXII03:CURRENT ELECTRICITY

357238 The reading of ideal voltmeter in the following circuit is
supporting img

1

3.4 V

2

5.0 V

3

2.4 V

4

4.0 V

Explanation:

As the voltmeter is ideal then given circuit is an open circuit, so reading of voltmeter is equal to the e.m.f. of cell, i.e., 5.0 $V$ .

PHXII03:CURRENT ELECTRICITY

357239 An ideal voltmeter and an ideal ammeter connected as shown in the figure. The reading of the voltmeter is
supporting img

1

6 V

2

9 V

3

3 V

4 Zero

Explanation:

Since ideal ammeter has zero resistance and potential drop across it, $V = 0$

PHXII03:CURRENT ELECTRICITY

357240 Read the Assertion and Reason carefully to mark the correct options given below:
Assertion :
For measuring the potential difference across a resistance of $600 Ω$ , the voltmeter with resistance $1000 Ω$ will be preferred over voltmeter with resistance $4000 Ω$ .
Reason :
Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.

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

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

3 Assertion is incorrect but Reason is correct.

4 Assertion is correct but reason is incorrect.

Explanation:

Assertion $A$ is incorrect because a voltmeter should have very high resistance. So that very small current flows through it.

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357241 Two batteries, one of emf $18 V$ and internal resistance $2 Ω$ , and the other of emf $12 V$ and internal resistance $1 Ω$ , are connected as shown. The voltmeter $V$ will record a reading of:
supporting img

1

15 V

2

30 V

3

14 V

4

18 V

Explanation:

Is clear that the two cells oppose each other hence, the effective emf in closed circuit is $18 - 12 = 6 V$ and net resistance is $1 + 2 = 3 Ω$ (because in the closed circuit the internal resistances of two cells are in series).
The current in circuit will be in direction of arrow shown in figure.
supporting img
$I = \frac{effective emf}{total resistance} = \frac{6}{3} = 2 A$
The potential difference across $V$ will be same as the terminal voltage of either cell.
Since, current is drawn from the cell of 18 volt, hence,
$\begin{aligned} V_{1} = E_{1} - i r_{1} \\ = 18 - (2 \times 2) = 18 - 4 = 14 V \end{aligned}$
Similarly, current enters in the cell of $12 V$ hence
$V_{2} = E_{2} + i r_{2}$
$= 12 + 2 \times 1 = 12 + 2 = 14 V$
Hence, $V = 14 V$

PHXII03:CURRENT ELECTRICITY

357242 A voltmeter with resistance $2500 Ω$ indicates a voltage of 125 $V$ in the circuit shown in figure. What is the series resistance to be connected to voltmeter so that it indicates 100 $V$ ? Consider ideal battery.
supporting img

1

625 Ω

2

120 Ω

3

550 Ω

4 Data are insufficient

Explanation:

As the battery is ideal emf should be 125 $V$ .
For second case: Let $R$ is the resistance to be connected in series with the voltmeter.
$(\frac{125}{2500 + R}) 2500 = 100 V$
$\Rightarrow R = 625 Ω$

PHXII03:CURRENT ELECTRICITY

357238 The reading of ideal voltmeter in the following circuit is
supporting img

1

3.4 V

2

5.0 V

3

2.4 V

4

4.0 V

Explanation:

As the voltmeter is ideal then given circuit is an open circuit, so reading of voltmeter is equal to the e.m.f. of cell, i.e., 5.0 $V$ .

PHXII03:CURRENT ELECTRICITY

357239 An ideal voltmeter and an ideal ammeter connected as shown in the figure. The reading of the voltmeter is
supporting img

1

6 V

2

9 V

3

3 V

4 Zero

Explanation:

Since ideal ammeter has zero resistance and potential drop across it, $V = 0$

PHXII03:CURRENT ELECTRICITY

357240 Read the Assertion and Reason carefully to mark the correct options given below:
Assertion :
For measuring the potential difference across a resistance of $600 Ω$ , the voltmeter with resistance $1000 Ω$ will be preferred over voltmeter with resistance $4000 Ω$ .
Reason :
Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.

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

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

3 Assertion is incorrect but Reason is correct.

4 Assertion is correct but reason is incorrect.

Explanation:

Assertion $A$ is incorrect because a voltmeter should have very high resistance. So that very small current flows through it.

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357241 Two batteries, one of emf $18 V$ and internal resistance $2 Ω$ , and the other of emf $12 V$ and internal resistance $1 Ω$ , are connected as shown. The voltmeter $V$ will record a reading of:
supporting img

1

15 V

2

30 V

3

14 V

4

18 V

Explanation:

Is clear that the two cells oppose each other hence, the effective emf in closed circuit is $18 - 12 = 6 V$ and net resistance is $1 + 2 = 3 Ω$ (because in the closed circuit the internal resistances of two cells are in series).
The current in circuit will be in direction of arrow shown in figure.
supporting img
$I = \frac{effective emf}{total resistance} = \frac{6}{3} = 2 A$
The potential difference across $V$ will be same as the terminal voltage of either cell.
Since, current is drawn from the cell of 18 volt, hence,
$\begin{aligned} V_{1} = E_{1} - i r_{1} \\ = 18 - (2 \times 2) = 18 - 4 = 14 V \end{aligned}$
Similarly, current enters in the cell of $12 V$ hence
$V_{2} = E_{2} + i r_{2}$
$= 12 + 2 \times 1 = 12 + 2 = 14 V$
Hence, $V = 14 V$

PHXII03:CURRENT ELECTRICITY

357242 A voltmeter with resistance $2500 Ω$ indicates a voltage of 125 $V$ in the circuit shown in figure. What is the series resistance to be connected to voltmeter so that it indicates 100 $V$ ? Consider ideal battery.
supporting img

1

625 Ω

2

120 Ω

3

550 Ω

4 Data are insufficient

Explanation:

As the battery is ideal emf should be 125 $V$ .
For second case: Let $R$ is the resistance to be connected in series with the voltmeter.
$(\frac{125}{2500 + R}) 2500 = 100 V$
$\Rightarrow R = 625 Ω$