3 Marks Question

Question 13 Marks

What do you mean by colour code of carbon resistance ?

Answer

Carbon resistors are too small in size to mark the value of their resistances on them. Therefore a colour code is used to indicate the resistance value and its percentage accuracy i.e. tolerence. The resistor has a set of four co-axial coloured rings of different colours on it.
First three rings (I, II, III) are made near one end of the cylinder and are close to each other. Fourth ring (IV) is made near the other end of the cylinder and its colour is either silver or gold. In carbon resistors of very small size all the four rings are very close to each other.

The colour of first two rings (bands) I, II, indicates the first two significant figures of the value of resistance of the resister in ohm. Third band (III) indicates the decimal multiplier i.e. the multiplier of power of 10. Fourth band (IV) indicates tolerance i.e. possible variation in percent about the indicated value of the resistance. Gold colour of IV band represents tolerance of ± 5% and silver colour of this band represents the tolerance of ± 10%, If fourth band is not made on the resister it means its tolerance will be $\pm 20$%.
$\therefore \quad$ Resistance $=\left[\right.$ I II × $1 0 ^{\text {III }} \pm$ IV $\left.\%\right]$

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Question 23 Marks

Explain that Kirchhoff's second law is law of conservation of energy.

Answer

According to Kirchhoff's second law :
$ \Sigma IR = \Sigma E $
On multiplying both sides by I. t, we get
$\sum I ^2 R t =\sum$ E.I. $t$
But $ \Sigma I^2Rt = $ Total energy consumed in different resistance present in the closed loop in time $t$.
$\sum$ E.I.$t$ Total energy given by the different sources of emfs i.e. cells present in the closed loop.
i.e. Total energy dissipiated = Total energy received.
This is according to law of conservation of energy.
In this way Kirchhoff's second law is the law of conservation of energy.

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Question 33 Marks

What is Peltier effect ?

Answer

Peltier Effect : In 1834 Peltier discovered an effect which is the converse of Seebeck effect. He found that when a current is passed across the junction of two dissimilar metals there is either absorption or evolution of heat at the junction, i.e., the junction is either cooled or heated. For a given pair of metals, the cooling or heating of the junction depends on the direction of the current. If a particular junction is heated by passing the current in one direction, the same is cooled when the direction of the current is reversed, i.e., Peltier effect is reversible.

To demonstrate Peliter effect two junctions A and B by joining two wire of copper and iron are obtained and current is passed as shown in adjoining fig. by joining a cell between A and B. We find, the junction at which the current enters from copper to iron is cooled and that at which current enters from iron to copper is heated.

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Question 43 Marks

When two resistors are connected in sereis and parallel, then their equivalent resistances are $ 16~\Omega $ and $ 3~\Omega $ respectively. Find out the resistance of each resistor.

Answer

Let the value of resistance of the two resistors be $ R_1 $ Ω and $ R_2 $ Ω. Equivalent resistance of series combination $ R_s = 16 $ and that of parallel combination $ R_p = 3 $Ω .
$\therefore$ $ R_1 + R_2 = R_s \Rightarrow R_1 + R_2 = 16~\Omega $ ...(1)
and $ \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_p} \Rightarrow \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3~\Omega} $ ...(2)
Now, from equation (2), we get
$ \frac{R_2 + R_1}{R_1 R_2} = \frac{1}{3~\Omega} $
⇒ $ R_1 R_2 = 3~\Omega \times (R_2 + R_1) $ ...(3)
On substituting the value of $\left( R _2+ R _1\right)$ from equation (1) in equation (3), we get
$R _1 R _2=3 \Omega \times 16 \Omega=48 \Omega^2$
But $ R_1 - R_2 = \sqrt{(R_1 + R_2)^2 - 4 R_1 R_2} $
$=\sqrt{(16)^2-4 \times 48}=\sqrt{64}$ ...(4)
$\therefore \quad R _1- R _2=8 \Omega$
On adding equations (1) and (4), we get
$2 R _1=24 \Omega \Rightarrow R _1=12 \Omega$
On substituting the value of $R _1$ in equation (1), we get
$12 \Omega+ R _2=16 \Omega \Rightarrow R _2=16 \Omega-12 \Omega \Rightarrow R _2= 4 \Omega$

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Question 53 Marks

What is specific conductivity ? Give its unit in S.I. system. Prove that $ \vec{j}=\sigma\vec{E} $, where $ \vec{E} = $ Intensity of electric field, $ \vec{j} = $ current density and $ \sigma = $ specific conductivity.

Answer

Specific Conductivity : The reciprocal of specific resistance of a material is called specific conductivity. It is denoted by $ \sigma $.
∴ Specific conductivity $=\frac{1}{\text { Specific Resistance }}$ i.e. $\sigma=\frac{1}{\rho}$
Proof of $ \vec{j} = \sigma \cdot \vec{E} $ : When an electric potential difference V is applied across the ends of an electric conductor of length l and area of cross-section A, then an electric field $ E = V/l $ is established inside the conductor. In this field, free electrons in the conductor begin to drift with drift velocity $ v_d $ due to which electric current I flowing through the conductor is given by: $ I = ne A v_d $ and
$ v_d = \frac{eV\tau}{ml} \Rightarrow v_d = e(\frac{V}{l})\frac{\tau}{m} \Rightarrow \frac{eE\tau}{m} $
$\therefore I = neA(\frac{eE\tau}{m}) = \frac{ne^2AE\tau}{m} $
$ \frac{I}{A} = (\frac{ne^2\tau}{m})E $
But $ \frac{I}{A} = current~density = j $
and $ \frac{m}{ne^2\tau} = \rho \Rightarrow \frac{ne^2\tau}{m} = \frac{1}{\rho} $
$ \Rightarrow j = (\frac{1}{\rho})E $ but $ \frac{1}{\rho} = \sigma $
$ j = \sigma E $ or in vector form $ \vec{j} = \sigma \cdot \vec{E} $

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