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