Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf Ωµ and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
(a) What is the value Ωµ ?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
(f ) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf Ωµ and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
(a) What is the value Ωµ ?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
(f ) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?
(a) Constant emf of the given standard cell, E1 = 1.02 V
Balance point on the wire, l1 = 67.3 cm
A cell of unknown emf, Ωµ,replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm
The relation connecting emf and balance point is,
The value of unknown emfis 1.247 V.
(b) The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.
(c) The balance point is not affected by the presence of high resistance.
(d) The point is not affected by the internal resistance of the driver cell.
(e) The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.
(f) The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.
The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
This question is important because it tests key concepts from the NCERT syllabus and is frequently asked in CBSE exams.
This question is based on core NCERT concepts explained in the chapter and should be revised thoroughly before exams.
Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.
(a) Constant emf of the given standard cell, E1 = 1.02 V
Balance point on the wire, l1 = 67.3 cm
A cell of unknown emf, Ωµ,replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 ...
Step-by-step explanation:
• (a) Constant emf of the given standard cell, E1 = 1
• 02 V
•
• Balance point on the wire, l1 = 67
• 3 cm
Stay updated with our latest educational content and study tips
Voice is a crucial aspect of English grammar. It indicates if the subject of the sentence is performing the action or if the action is performed on the subject. To know the difference between the active and passive voice, contributing to the improvement of writing, speaking, and sentence clarity. In this article, you will learn … Read more
Read More
Have you ever noticed that doing activities such as playing outdoors, chilling with your friends, or using the mobile phone feels so effortless and fun? Whenever you might have played sports, you kept on playing even when it felt hard initially just so you could get that feeling after the game, a feeling of accomplishment, … Read more
Read More
Imagine this; You’re sitting in the test hall, pen in hand, and the question paper has been put in front of you. Rather than freezing, you feel calm, collected, and set. Why? As you have formerly been there in your mind. You’ve seen yourself sitting composibly, reviewing the question paper fluently, performing well in the … Read more
Read More
Parent-Teacher Meetings (PTMs) are more than quick updates on marks — they’re a chance to build a real partnership between home and school. A good Parent-Teacher Meeting conversation helps parents see beyond grades. It opens up insights about a child’s strengths, struggles, emotions and even hidden talents. When parents participate actively, they don’t just track … Read more
Read More
Student Discussion
ADD NEW COMMENT