3.22 Figure 3.33 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 and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
(a) What is the value e ?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?
(e) 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 cell of unknown emf, , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm.
The relation of connected emf and balance point is, =
Hence, = = = 1.247 V
The purpose of using high resistance of 600 KΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.
The balance point is not affected by the presence of high resistance.
The point is not affected by the internal resistance of the driver cell.
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3.22 Constant emf of the standard cell, = 1.02 V
Balance point on the wire, = 67.3 cm
A cell of unknown emf, , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm.
The relation of connected emf and balance point is, =
Hence, = = = 1.247 V
The purpose of using high resistance of 600 KΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.
The balance point is not affected by the presence of high resistance.
The point is not affected by the internal resistance of the driver cell.
The method would not work if the emf of 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.
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.
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<p><strong>3.22 </strong>Constant emf of the standard cell, <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span> = 1.02 V</p><p>Balance point on the wire, <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>l</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span> = 67.3 cm</p><p>A cell of unknown emf, <span title="Click to copy mathml"><math><mi>?</mi></math></span> , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm.</p><p>The relation of connected emf and balance point is, <span title="Click to copy mathml"><math><mfrac><mrow><mrow><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>l</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></mrow></mrow></mfrac></math></span> = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>?</mi></mrow></mrow><mrow><mrow><mi>l</mi></mrow></mrow></mfrac></math></span></p><p>Hence, <span title="Click to copy mathml"><math><mi>?</mi><mi></mi></math></span> = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>l</mi></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>l</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></mrow></mrow></mfrac><mo>×</mo><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span> = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>82.3</mn></mrow></mrow><mrow><mrow><mn>67.3</mn></mrow></mrow></mfrac><mo>×</mo><mn>1.02</mn></math></span> = 1.247 V</p><p>The purpose of using high resistance of 600 KΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.</p><p>The balance point is not affected by the presence of high resistance.</p><p>The point is not affected by the internal resistance of the driver cell.</p><p>The method would not work if the emf of 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.</p><p>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.</p><p>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.</p>
According to this chapter, a galvanometer is used to find and measure the small electric currents in a circuit. The principle that works in a galvanometer is the electromagnetic induction.
There are two types of electricity - Static and Current electricity. The electric charges buildup on a material's surface is called the static electricity. The continuous flow of electric charge is termed as the current electricity. Current electricity is of two types - Alternating Current (AC) and Direct Current (DC). In AC, the charge direction reverses periodically and in DC, charge flows in one direction.
In simple words, current electricity can be defined as the electric charge continuously moving from one place to another along a pathway. It is measured in amperes (A). Electric current is needed for electrical devices to work.
No, in fact, it is one of the easiest chapter of class 12 Physics. Other chapters which are considered comparatively easy are Ray Optics and Electric Charges and Fields.
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