2. Solved Papers
  3. Uttarakhand PMT

Uttarakhand PMT Uttarakhand PMT Solved Paper-2010

  • question_answer
    In a noiseless transformer, an alternating current of 2 A is flowing in the primary coil. The number of turns on the primary and secondary coils are 100 and 20 respectively. The value of the current in the secondary coil is

    A)  0.08 A          

    B)  0.4 A

    C)  5 A            

    D)  10 A

    Correct Answer: D

    Solution :

     \[\frac{{{N}_{s}}}{{{N}_{p}}}=\frac{{{I}_{p}}}{{{I}_{s}}}\Rightarrow \frac{20}{100}=\frac{2}{{{I}_{s}}}\] \[=\frac{1}{5}=\frac{2}{{{I}_{s}}}\] \[\therefore \]The current in the secondary coil \[{{I}_{s}}=5\times 2=10A\]


You need to login to perform this action.
You will be redirected in 3 sec spinner