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What conclusion can you draw from the follwing observation on a resistor made of alloy manganin ? |
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Answer» SOLUTION :(1) ` R = (V)/(I) = (3.94)/(0.2) = 19.7 Omega` (2) `R = (V)/(I) = (7.87)/(0.4) = 19.67 Omega` (3) ` R = (V)/(I) = (11.8)/(0.6) = 19.66 Omega` (4) `R= (V)/(I)(15.7)/(0.8) = 19.62 Omega` `(5)R = (V)/(I)= (19.7)/(1.0) = 19.7 Omega ` `(6)R= (V)/(I) = (39.4)/(2.0) = 19.7 Omega` `(7) R= (V)/(I) = (59.2)/(3.0) = 19.7 Omega` `(8) R = (V)/(I) = (78.8)/(4.0) = 19.7 Omega` `(9) R= (V)/(I) = (98.5)/(5.0) = 19.72 Omega` (10) ` R = (V)/(I) = (118.5)/(6.0) = 19.75 Omega` (11) `R = (V)/(I)=(138.2 )/(7.0) = 19.74 Omega` (12) ` R = (V)/(I) = (158.0)/(8.0) = 19.75 Omega` From above calculations we can conclude that for GIVEN alloy, its resistance R = `(V)/(I) `remains constant and hence obeys Ohm.s law. |
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