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State the principle of a transformer. Explain its construction and working. Derive an expression for the ratio of e.m.f's in terms of number of turns in primary and secondary coil. Two diametrically opposite pointsof a metal ring are connected to two terminals of the left gap of meter bridge. The resistance of 11Omega is connected in right gap. If null point is obtained at a distance of 45 cm from the left end, find the resistance of metal ring. |
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Answer» Solution :Numerical : Given : `R_(2)=11Omega,l_(1)=45" cm",l_(2)=100-45=55" cm"` RESISTANCE of METAL ring `(R_(1))=?` `(R_(1))/(R_(2))=(l_(1))/(l_(2))` Resistance of each half segment of the metal ring `=(R_(1))/(2)`. These half segments are connected in parallel in the left gap. `R_(eff.)=((R_(1))/(2)xx(R_(1))/(2))/((R_(1))/(2)+(R_(1))/(2))=((R_(1))/(2)xx(R_(1))/(2))/(R_(1))=(R_(1)^(2))/(4R_(1))` `R_(eff.)=(R_(1))/(4)OMEGA` From the formula, `(R_(eff.))/(R_(2))=(l_(1))/(l_(2))` `((R_(1))/(4))/(11)=(45)/(55)` `(R_(1))/(44)=(45)/(55)` `R_(1)=(45)/(55)xx44` `R_(1)=36Omega` `:.` The resistance of metal ring is `36Omega`. |
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