question_answer

A transformer is having 2100 turns in primary and 4200 turns in secondary. An AC source of 120 V, 10 A is connected to its primary. The secondary voltage and current are :

A)  240 V, 5 A     

B)  120 V, 10 A

C)  240 V, 10 A   

D)  120 V, 20 A

Correct Answer: A

Solution :

 Let \[{{i}_{p}}\] and \[{{i}_{S}}\] be the currents in primary and secondary at any instant and energy losses be zero, then power in secondary = power in primary \[{{V}_{S}}\times {{i}_{S}}={{V}_{p}}\times {{i}_{p}}\] \[\frac{{{i}_{P}}}{{{i}_{S}}}=\frac{{{V}_{S}}}{{{V}_{P}}}=\frac{{{N}_{S}}}{{{N}_{P}}}=r\] \[\therefore \] \[{{V}_{S}}=\frac{{{N}_{S}}}{{{N}_{P}}}\times {{V}_{P}}\] Given, \[{{V}_{P}}=120\,V,\,{{N}_{S}}=4200,\,{{N}_{P}}=2100\] \[\therefore \] \[{{V}_{S}}=\frac{4200}{2100}\times 120\] \[{{V}_{S}}=240\,V\] and \[\frac{{{I}_{S}}}{{{I}_{P}}}=\frac{{{N}_{P}}}{{{N}_{S}}}\] \[\Rightarrow \] \[{{I}_{S}}=\frac{{{N}_{P}}}{{{N}_{S}}}\times {{I}_{P}}\] \[=\frac{2100}{4200}\times 10=5\,A\]. Note: When voltage is stepped up, the current is correspondingly reduced in the same ratio, and wee versa. Thus, energy obtained from secondary coil is equal to the energy given to the primary coil.