A) 15 A
B) 50 A
C) 25 A
D) 12.5 A
Correct Answer: B
Solution :
Key Idea: The flux per turn of primary coil must be equal to flux per turn of the secondary coil. As per key idea, \[\frac{{{\phi }_{p}}}{{{n}_{p}}}=\frac{{{\phi }_{s}}}{{{n}_{s}}}\] or \[\frac{1}{{{n}_{p}}}.\frac{d{{\phi }_{p}}}{dt}=\frac{1}{{{n}_{s}}}\frac{d{{\phi }_{s}}}{dt}\] \[\therefore \frac{{{e}_{s}}}{{{e}_{p}}}=\frac{{{n}_{s}}}{{{n}_{p}}}\left( as\,\,\,e\propto \,\frac{d\phi }{dt} \right)\] For no loss of power, ei = constant \[\therefore i=\frac{1}{e}\times \text{constant}\] or \[\frac{{{i}_{p}}}{{{i}_{s}}}=\frac{{{e}_{s}}}{{{e}_{p}}}\] or \[\frac{{{i}_{p}}}{{{i}_{s}}}=\frac{{{n}_{s}}}{{{n}_{p}}}\] Here, \[\frac{{{n}_{p}}}{{{n}_{s}}}=\frac{1}{25},\,{{i}_{s}}=2A\] \[\therefore \frac{{{i}_{p}}}{2}=\frac{25}{1}\] or \[{{i}_{p}}=25\times 2=50\,A\] Note: In step-up transformer \[{{n}_{s}}>{{n}_{p}}\]. It increases voltage and reduces current. In step-down transformer, \[{{n}_{p}}>{{n}_{s}}\]. It increases current and reduces voltage.
You need to login to perform this action.
You will be redirected in
3 sec
