
State the principle on which transformer works. Explain its working with construction. Derive an expression for ration of e.m.f.s and currents in terms of number of turns in primary and secondary coil.
A conductor of any shape, having area $40c{m^2}$ placed in an air is uniformly charged with a charge $0.2\mu C$. Determine the electric intensity at a point just outside its surface. Also find the mechanical force per unit area of the charged conductor. $\left( {{ \in _0} = 8.85 \times {{10}^{ - 12}}S.I.units} \right)$.
Answer
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Hint:The Principle of working of a transformer is that whenever the magnetic flux linked with a coil changes, an emf is induced in the neighboring coil.The idea of a transformer was first discussed by Michael Faraday in the year 1831 and was carried forward by many other prominent scientific scholars. However, the general purpose of using transformers was to maintain a balance between the electricity that was generated at very high voltages and consumption which was done at very low voltages.
Formula Used:
The electric intensity at a point just outside its surface is:
$E = \dfrac{\sigma }{{{\varepsilon _0}}}$
The mechanical force per unit area of the charged conductor is:
$f = \dfrac{1}{2}{\varepsilon _0}{E^2}$
Complete step by step answer:
The Principle of working of a transformer is that at whatever point the attractive motion connected with a loop changes, an emf is incited in the neighboring curl. The development of a transformer is portrayed as follows: It comprises two loops essential (P) and optional (S). These two curls are protected from one another. Likewise, they are twisted on a delicate iron center. The essential loop is otherwise called the information curl and the auxiliary curl is otherwise called the yield loop.
The working of a transformer is portrayed as follows: Whenever a rotating voltage is applied to the essential curl, the current through the loop changes. Furthermore, accordingly, the attractive motion through the center likewise changes. An emf is actuated in every one of them as this changing attractive transition is connected with the two loops. The measure of attractive motion connected with the loop relies upon the quantity of turns the curl takes.
Derivation: Let be the magnetic flux, linked per turn with both coils at a certain instant of time $f$.Let the number of turns of the primary and secondary coils be ${N_p}$and ${N_s}$ respectively. Therefore, the total magnetic flux linked with the primary coil can be written as ${N_p}\varphi $. Similarly, the total magnetic flux linked with the secondary coil is ${N_s}\varphi $.Induced emf in a given coil is,
$e = \dfrac{{d\phi }}{{dt}}$
This formula can be used to write induced emf in primary and secondary coils.
$e{_p} = - \dfrac{{d\phi p}}{{dt}} \\
\Rightarrow e{_p} = - \dfrac{{dNp\phi }}{{dt}} \\
\Rightarrow e{_p} = - Np\dfrac{{d\phi }}{{dt}}$ $ \to (1)$
$\Rightarrow e{_s} = - \dfrac{{d\phi s}}{{dt}} \\
\Rightarrow e{_s} = - \dfrac{{dNs\phi }}{{dt}} \\
\Rightarrow e{_s} = - Ns\dfrac{{d\phi }}{{dt}}$ $ \to (2)$
Dividing equations (1) and (2)
$\dfrac{{{e_s}}}{{{e_p}}} = \dfrac{{ - Ns\dfrac{{d\phi }}{{dt}}}}{{ - N\dfrac{{d\phi }}{{dt}}}} \\
\therefore\dfrac{{{e_s}}}{{{e_p}}}= \dfrac{{Ns}}{{Np}}\;$ $ \to (3)$
Turns ratio of the transformer is given by this equation.
Numerical: It is given that area is $A = 40c{m^2} = 4 \times {10^{ - 6}}C$
charge is$Q = 0.2\mu C = 0.2 \times {10^{ - 6}}C$
The electric intensity at a point just outside its surface is:
$E = \dfrac{\sigma }{{{\varepsilon _0}}} = \dfrac{Q}{{A{\varepsilon _0}}}$
$\Rightarrow E = \dfrac{{0.2 \times {{10}^{ - 6}}}}{{40 \times {{10}^{ - 4}} \times 8.85 \times {{10}^{ - 12}}}} \\
\therefore E= 5.56 \times {10^6}N/C$
The mechanical force per unit area of the charged conductor is
$f = \dfrac{1}{2}{\varepsilon _0}{E^2} \\
\Rightarrow f= \dfrac{1}{2} \times 8.85 \times {10^{ - 12}} \times {\left( {5.56 \times {{10}^6}} \right)^2} \\
\therefore f= 12141.25N/{m^2}$
Note:The total magnetic flux linked with the primary and the secondary coil is given at a particular interval of time. Note that the formula for electric field just outside the surface, that we used in the above numerical is only valid for a charged conductor. This is because the electric field vectors at the surface (outside) of a conductor are always perpendicular to the surface and we can consider the surface as an infinitely large planar charged body.
Formula Used:
The electric intensity at a point just outside its surface is:
$E = \dfrac{\sigma }{{{\varepsilon _0}}}$
The mechanical force per unit area of the charged conductor is:
$f = \dfrac{1}{2}{\varepsilon _0}{E^2}$
Complete step by step answer:
The Principle of working of a transformer is that at whatever point the attractive motion connected with a loop changes, an emf is incited in the neighboring curl. The development of a transformer is portrayed as follows: It comprises two loops essential (P) and optional (S). These two curls are protected from one another. Likewise, they are twisted on a delicate iron center. The essential loop is otherwise called the information curl and the auxiliary curl is otherwise called the yield loop.
The working of a transformer is portrayed as follows: Whenever a rotating voltage is applied to the essential curl, the current through the loop changes. Furthermore, accordingly, the attractive motion through the center likewise changes. An emf is actuated in every one of them as this changing attractive transition is connected with the two loops. The measure of attractive motion connected with the loop relies upon the quantity of turns the curl takes.
Derivation: Let be the magnetic flux, linked per turn with both coils at a certain instant of time $f$.Let the number of turns of the primary and secondary coils be ${N_p}$and ${N_s}$ respectively. Therefore, the total magnetic flux linked with the primary coil can be written as ${N_p}\varphi $. Similarly, the total magnetic flux linked with the secondary coil is ${N_s}\varphi $.Induced emf in a given coil is,
$e = \dfrac{{d\phi }}{{dt}}$
This formula can be used to write induced emf in primary and secondary coils.
$e{_p} = - \dfrac{{d\phi p}}{{dt}} \\
\Rightarrow e{_p} = - \dfrac{{dNp\phi }}{{dt}} \\
\Rightarrow e{_p} = - Np\dfrac{{d\phi }}{{dt}}$ $ \to (1)$
$\Rightarrow e{_s} = - \dfrac{{d\phi s}}{{dt}} \\
\Rightarrow e{_s} = - \dfrac{{dNs\phi }}{{dt}} \\
\Rightarrow e{_s} = - Ns\dfrac{{d\phi }}{{dt}}$ $ \to (2)$
Dividing equations (1) and (2)
$\dfrac{{{e_s}}}{{{e_p}}} = \dfrac{{ - Ns\dfrac{{d\phi }}{{dt}}}}{{ - N\dfrac{{d\phi }}{{dt}}}} \\
\therefore\dfrac{{{e_s}}}{{{e_p}}}= \dfrac{{Ns}}{{Np}}\;$ $ \to (3)$
Turns ratio of the transformer is given by this equation.
Numerical: It is given that area is $A = 40c{m^2} = 4 \times {10^{ - 6}}C$
charge is$Q = 0.2\mu C = 0.2 \times {10^{ - 6}}C$
The electric intensity at a point just outside its surface is:
$E = \dfrac{\sigma }{{{\varepsilon _0}}} = \dfrac{Q}{{A{\varepsilon _0}}}$
$\Rightarrow E = \dfrac{{0.2 \times {{10}^{ - 6}}}}{{40 \times {{10}^{ - 4}} \times 8.85 \times {{10}^{ - 12}}}} \\
\therefore E= 5.56 \times {10^6}N/C$
The mechanical force per unit area of the charged conductor is
$f = \dfrac{1}{2}{\varepsilon _0}{E^2} \\
\Rightarrow f= \dfrac{1}{2} \times 8.85 \times {10^{ - 12}} \times {\left( {5.56 \times {{10}^6}} \right)^2} \\
\therefore f= 12141.25N/{m^2}$
Note:The total magnetic flux linked with the primary and the secondary coil is given at a particular interval of time. Note that the formula for electric field just outside the surface, that we used in the above numerical is only valid for a charged conductor. This is because the electric field vectors at the surface (outside) of a conductor are always perpendicular to the surface and we can consider the surface as an infinitely large planar charged body.
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