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A short-circuited coil is placed in a time varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the radius of the wire is to be halved, then find the electrical power dissipated |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> : Current is induced in the short-circuited coil due to the imposed time - varying magnetic field. <br/>Power P = `(e^2)/(R )` <br/>Here e ` = - (dphi)/(dt)` where `<a href="https://interviewquestions.tuteehub.com/tag/phi-599602" style="font-weight:bold;" target="_blank" title="Click to know more about PHI">PHI</a> = NBA` <br/>and `R = (<a href="https://interviewquestions.tuteehub.com/tag/rho-623364" style="font-weight:bold;" target="_blank" title="Click to know more about RHO">RHO</a> l)/(pi r^2) ` where l and r are <a href="https://interviewquestions.tuteehub.com/tag/length-1071524" style="font-weight:bold;" target="_blank" title="Click to know more about LENGTH">LENGTH</a> and radius of the wire<br/>` therefore P = (pi r^2)/(rho l) [ (d)/(dl) (NBA) ]^2 " or " P = (pi r^2)/(rho l) N^2 A^2 ((dB)/(dt))^2`<br/>or P = (constant) `(N^2r^2)/(l)` <br/>When `r_2= (r_1)/(2)` then `l_2 = 4l_1` <br/>` P_2/P_1 = ((4N)^2)/(N^2) xx ((r)/(2r))^2 xx ((l)/(4l))`<br/>` therefore P_2/P_1 = (16N^2 xx r^2 xx l)/(N^2 xx 4r^2 xx 4l) " or " P_2/P_1 = 1/1 `<br/>`therefore ` Power dissipated is the same</body></html> | |