Why is voltmeter better than a potentiometer for measuring EMF of cell?

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A voltmeter and a potentiometer are both used for measuring the emf of a cell. However, a voltmeter is generally considered better for this purpose due to the following reasons:

Accuracy: A voltmeter provides a more accurate reading than a potentiometer. A voltmeter measures the emf directly, while a potentiometer measures the potential difference across a known resistor and then calculates the emf. Any errors or variations in the resistance of the known resistor can affect the accuracy of the reading.
Ease of use: A voltmeter is easier to use than a potentiometer. A voltmeter can be connected directly across the terminals of the cell and the reading can be obtained immediately. On the other hand, a potentiometer requires a more complex setup involving a known resistor and a sliding contact to obtain the emf reading.
Time-saving: A voltmeter is quicker than a potentiometer. A voltmeter provides an instant reading, while a potentiometer requires time for setting up the circuit and taking the measurement.
Cost-effective: A voltmeter is more cost-effective than a potentiometer. A basic voltmeter is cheaper than a potentiometer, which requires a more complex setup involving a known resistor and a sliding contact.

Hence, due to its accuracy, ease of use, time-saving, and cost-effectiveness, a voltmeter is considered better than a potentiometer for measuring the emf of a cell.

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1.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

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2.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

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3.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

View Solution

4.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

5 V
3.5 V
2.5 V
Zero V

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5.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

(i) and (iii) only
(iii) and (iv) only
(i), (iii) and (iv) only
All of them

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6.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Both Assertion (A) and Reason (R) are false.

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Why is voltmeter better than a potentiometer for measuring emf of cell?

Related Questions

CBSE CLASS XII Related Questions

1.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

2.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

3.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

4.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

5.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

6.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

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Why is voltmeter better than a potentiometer for measuring emf of cell?

Related Questions

CBSE CLASS XII Related Questions

1.A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

2.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

3. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

4.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

5.Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

6.Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

2.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

3.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

4.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

5.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

6.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.