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What is the maximum voltage (emf) created by a 50 loop de generator with a cross-sectional area of 0.5 square meter if the ma
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Answer #1

Sol:

Here N = 50 A=0.5m^2  \Delta B=2T and \Delta t=0.2s

Hence induced emf is-

\varepsilon =N\left ( \frac{A\Delta B}{\Delta t} \right )=50\times \left ( \frac{0.5\times 2}{0.2} \right )=250V

Here N = 100 A=0.1m^2  B=0.5T and f = 60Hz

Maximum induced emf is-

\varepsilon _{max}=2\pi fNBA=2\pi\times60 \times 100\times0.5\times0.1=1885V

Here N1 = 200 V1 = 120V and V2 = 12 V

Now turn in secondary coil-

N_{2}=\frac{V_{2}}{V_{1}}\times N_{1}=\frac{12}{120}\times 200=20 turn

Now I1 =4A then

I_{2}=\frac{I_{1}V_{1}}{V_{2}}=\frac{4\times 120}{12}=40A

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