A) 50 watt
B) 54 watt
C) 104 watt
D) 4 watt
Correct Answer: B
Solution :
[b] \[{{E}_{c}}=100\,V,\,\,\,{{m}_{a}}=0.4,\,R=100\Omega ,\] \[{{P}_{c}}=\frac{{{E}_{c}}^{2}}{2R}=\frac{{{(100)}^{2}}}{2\times 100}=50\] watt \[P=\left( 1+\frac{{{m}_{a}}^{2}}{2} \right){{P}_{c}}=\left[ 1+\frac{{{(0.4)}^{2}}}{2} \right]\times 50=54\]watt
You need to login to perform this action.
You will be redirected in
3 sec
