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Generator Quiz 2

physics

If the motor of a generator has just started moving, then what will be the armature current? 

When the armature has just started moving there is no back emf, i.e. voltage that is opposite to the current flow, so armature current is equal to its maximum value. The total emf is given by ${{V}_{total}}=V-{{V}_{b}}$.Here, $V$ is the initial emf and ${{V}_{b}}$ is the back emf, ${{V}_{b}}=0$, as it depends on the angular frequency of the coil. 

 Generator is a device that converts mechanical energy into electrical energy, they work on the principle of Faraday’s law of electromagnetic induction, the armature of the generator rotates to produce a current in the coil. When the armature has just started moving there is no back emf, i.e. voltage which is opposite to the current flow, so armature current is equal to its maximum value. The total emf is given by, $ \\ {{V}_{total}}=V-{{V}_{b}} \\ $ So, if when the armature has just started moving, ${{V}_{b}}=0$, as it depends on the angular frequency of the coil, that is minimum at the start of the motor, therefore, ${{V}_{total}}$ is maximum. 

 Consider an AC generator, the area of the coil is $1\,{{m}^{2}}$ and number of winding number of the coil is $250$. If the coil rotates in a magnetic field strength of $3$ W/${{m}^{2}}$ at the frequency of $300$ rotations/sec, then what is the peak emf induced? 

The emf induced in a coil due to the magnetic field $B$, is the rate of change of magnetic flux linked with the coil and it depends on the number of turns of the coil $N$, and the area of cross-section $A$ and the angular frequency $\omega $ ,so emf is given by $\epsilon =NBA\omega =NBA\times 2\pi f$. Put in the known values given in the question to calculate the emf induced in the coil. 

 Generators are devices that convert mechanical energy into electrical energy, they work on the principle of Faraday’s law of electromagnetic induction. The induced emf due to the magnetic field is given by, $ \\ \epsilon =NBA\omega =NBA\times 2\pi f ----(i) \\ $ According to the question, $ \\ N=250,A=1{{m}^{2}},B=3$ W/${{m}^{2}}$ and $f=300$ revolutions /sec. $ \\ $ Putting the above values in equation (i), we get, $ \\ \epsilon =NBA\times 2\pi f \\ \epsilon =250\times 3\times 1\times 2\pi \times 300 \\ \therefore\epsilon =~1.4\times {{10}^{6}}V~$ 

How much should the magnetic field strength be changed to produce an emf equal to $565.2\,kV$ using a generator with coil of area $3\,{{m}^{2}}$ wound $100$times, having frequency $50$ rotations/second and the magnetic field strength as $5$T? 

The emf induced in a coil due to the magnetic field $B$, is the rate of change of magnetic flux linked with the coil and it depends on the number of turns of the coil $N$, and the area of cross-section $A$ and the angular frequency $\omega $ ,so emf is given by $\epsilon =NBA\omega =NBA\times 2\pi f$.Put in the known values given in the question to calculate the magnetic field in both the cases to find the percentage increase in magnetic field. 

We have to calculate the percentage increase, so let the initial magnetic field strength be ${{B}_{1}}$ and the final magnetic field strength be ${{B}_{2}}$. The induced emf due to the magnetic field is given by, $ \\ \epsilon =NBA\omega =NBA\times 2\pi f  ----(i) \\ $ According to the question, initially, $ \\ N=100,A=3{{m}^{2}},{{B}_{1}}=5T$ and $f=50$ revolutions /sec $ \\ $ We have to find the magnetic field which induces an emf of $565.2kV$, so putting the above values in equation (i), we get, $ \\ \epsilon =N{{B}_{2}}A\times 2\pi f \\ \Rightarrow {{B}_{2}}=\dfrac{\epsilon }{NA\times 2\pi f} \\  \Rightarrow {{B}_{2}} =\dfrac{565.2\times 1000}{100\times 3\times 2\pi \times 50} \\  \Rightarrow {{B}_{2}} =6.0T \\ $ Let $x$ be the percentage increase in the original magnetic field, then, $ \\ {{B}_{2}}={{B}_{1}}+\left({{B}_{1}}\times \dfrac{x}{100}\right) \\ \Rightarrow 6=5+\left(5\times \dfrac{x}{100}\right) \\ \therefore x=20\%$  

The slip rings and brushes in an AC generator, 

Slip rings and brushes in an AC generator are the devices that are used to connect the stationary components and rotating components of the generator to transmit power. 

An AC generator is a device that converts mechanical energy into electrical energy, they work on the principle of Faraday’s law of electromagnetic induction. A generator consists of a coil that is placed in a magnetic field and it is made to rotate, as a result the magnetic flux linked with the coil changes which in turn induces an emf in the coil. Slip rings and brushes in an AC generator are the devices that are used to connect the stationary components and rotating components of the generator to transmit power. The slip rings that are connected to the armature coil provide electric contact with the brushes. 

How many times should the coil of area $0.5\,{{m}^{2}}$ be wound in a AC generator to produce an emf equal to $879.2\,kV$, if it is placed in a magnetic field of $4$ W/${{m}^{2}}$ and the coil rotates $12000$ rotations/min? 

The emf induced in a coil due to the magnetic field $B$, is the rate of change of magnetic flux linked with the coil and it depends on the number of turns of the coil $N$, and the area of cross-section $A$ and the angular frequency $\omega $ ,so emf is given by $\epsilon =NBA\omega =NBA\times 2\pi f$. Put in the known values in the above equation and cross-multiply to get a number of turns. 

Generators are devices that convert mechanical energy into electrical energy, they work on the principle of Faraday’s law of electromagnetic induction. The induced emf due to the magnetic field is given by, $ \\ \epsilon =NBA\omega =NBA\times 2\pi f  ----(i) \\ $ According to the question, $ \\ A=0.5{{m}^{2}},\epsilon =879.2kV=879.2\times {{10}^{3}}V,B=4W/{{m}^{2}}$ and $f=12000$ revolutions /sec. $ \\ $ Putting the above values in equation (i) and solving for B, we get, $ \\ N=\dfrac{\epsilon }{2\pi BAf} \\ \Rightarrow N =\dfrac{879.2\times {{10}^{3}}}{2\pi \times 4\times 0.5\times 12000} \\ \therefore N =350$ 