A) \[\frac{{{\left( 10 \right)}^{4}}}{25{{\left( 8 \right)}^{2}}}V\]
B) \[\frac{{{10}^{4}}}{\left( 25 \right)\left( 8 \right)}V\]
C) \[\frac{{{\left( 10 \right)}^{4}}\,8}{25}V\]
D) \[\frac{{{\left( 10 \right)}^{4}}\,{{\left( 8 \right)}^{2}}}{25}V\]
Correct Answer: C
Solution :
Transformer works on the principle of mutual induction. Power in the secondary = Power in the primary \[{{V}_{s}}\times {{i}_{s}}={{V}_{p}}\times {{i}_{p}}\] Where \[{{i}_{s}}\],\[{{i}_{p}}\] are currents in the secondary and primary and \[{{V}_{s}}\] and \[{{V}_{p}}\] are voltages across secondary and primary respectively. Also, \[\frac{{{i}_{p}}}{{{i}_{s}}}=\frac{{{V}_{s}}}{{{V}_{p}}}=\frac{{{N}_{s}}}{{{N}_{p}}}=r=\] transformation ratio Given, \[{{P}_{p}}=10\,kW=10\times {{10}^{3}}\,\,W={{10}^{4}}\,W\] \[{{i}_{s}}=25\,A\] \[{{N}_{p}}:{{N}_{s}}=8:1\] Then, \[\frac{{{i}_{s}}}{{{i}_{p}}}=\frac{{{N}_{p}}}{{{N}_{s}}}=\frac{8}{1}\] \[\Rightarrow \] \[{{i}_{p}}=\frac{{{i}_{s}}}{8}=\frac{25}{8}A.\] Also, \[{{10}^{4}}={{V}_{p}}{{i}_{p}}\] \[\left( P=Vi \right)\] \[\Rightarrow \] \[{{V}_{p}}=\frac{{{10}^{4}}}{25/8}={{10}^{4}}\left( \frac{8}{25} \right)V\]
You need to login to perform this action.
You will be redirected in
3 sec
