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Published on: 18/06/2021
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Questions + Answers key
Take MCQ Physics Test1.
Explain Peltier effect.
2.
Find the expression for the equivalent emf & internal resistance of the series combination of cells.
3.
Derive a relation between internal resisance and emf of a cell.
4.
(a) Distinguish between electric cells and batteries.
(b) Explain its function.
5.
Derive an expression of drift velocity and write the relation between drift velocity and mobility.
1.
(i) When an electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and absorbed at the other junction. This is known as the Peltier effect.
(ii) In the Cu-Fe thermocouple the junctions A and B are maintained at the same temperature.
(iii) Let a current from a battery flow through the thermocouple. At junction A, where the current flows from Cu to Fe, heat is absorbed and junction A becomes cold.
(iv) At junction B, where the current flows from Fe to Cu heat is liberated and it becomes hot.

Peltier effect: Cu - Fe thermocouple
(v) When the direction of current is reversed, junction A gets heated and junction B gets cooled as shown in Figure (b).
Hence Peltier effect is reversible.
2.
(i) Suppose n cells, each of emf \(\xi \) volts and internal resistance r ohms are connected in series with an external resistance R.
(ii) The total emf of the battery = \(n\xi \) The total resistance in the circuit = nr + R By Ohm's law, the current in the circuit is
\(I=\cfrac { totalemf }{ taoal\ resistance } =\cfrac { n\xi }{ nr+R } \)
\(I=\cfrac { n\xi }{ R } =n{ l }_{ 1 }\)
(iii) where II is the current due to a single cell
\(\left( { I }_{ 1 }=\cfrac { \xi }{ R } \right) \)
Thus, if r is negligible when compared to R the current supplied by the battery is n times that supplied by a single cell.
Case (b) If r >> R,\(I=\cfrac { n\xi }{ nr } =\cfrac { \xi }{ r } \)
(iv) It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells.
(v) Thus series connection of cells is advantageous only when the effective internal resistance of the cells is negligibly small compared with R.
3.
(i) The emf of cell \(\xi \) is measured by connecting a high resistance voltmeter across it without connecting the external resistance R.

(ii) Since the voltmeter draws very little current for deflection, the circuit may be considered as open. Hence the voltmeter reading gives the emf of the cell.
(iii) Then, external resistance R is included in the circuit, and current I is established in the circuit. The potential difference across R is equal to the potential difference across the cell (V).
(iv) The potential drop across the resistor R is V + IR
(v) Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell . It is because a certain amount of voltage (Ir) has dropped across the internal resistance r.
Then \(V=\xi -Ir\)
\(Ir=\xi -V\)
(vi) Dividing equation (2) by equation (1) we get
\(\cfrac { Ir }{ IR } =\cfrac { \xi -V }{ V } \)
\(r=\left| \cfrac { \xi -V }{ V } \right| R\)
Since \(\xi \) V and R are known, internal resistance r can be determined.
4.
(a) An electric cell converts chemical energy into electrical energy to produce electricity. It contains two electrodes immersed in an electrolyte as shown Several electric cells connected together form a battery.

(b) When a cell or battery is connected to a circuit, electrons flow from the negative terminal to the positive terminal through the circuit. By using chemical reactions, a battery produces potential differences across its terminals. This potential difference provides the energy to move the electrons through the circuit.
5.
The drift velocity is the average velocity acquired by the electrons inside the conductor when it is subjected to an electric field. The average time between successive collisions is called the mean free time denoted by \(\tau \). The acceleration \(\vec { a } \) experienced by the electron in an electric field \(\vec { E } \) is given by
\(\vec { a } =\cfrac { -e\vec { E } }{ m } \left( since\vec { F } =-e\vec { E } \right) \)
The drift velocity is given by
\({ \vec { V } }_{ d }=\vec { a } \tau \)
\({ \vec { V } }_{ d }=\cfrac { e\tau }{ m } \vec { E } \)
\({ \vec { V } }_{ d }=-\mu \vec { E } \)
Here \(\mu =\cfrac { e\tau }{ m } \) is the mobility of the electron and it is defined as the magnitude of the drift velocity per unit electric field \(\mu =\cfrac { \left| { \vec { v } }_{ d } \right| }{ \left| \vec { E } \right| } \)
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