A) 50%
B) 60%
C) 65%
D) 71%
Correct Answer: D
Solution :
We know that \[{{\left( \frac{{{I}_{t}}}{{{I}_{c}}} \right)}^{2}}=1+\frac{{{m}^{2}}}{2}\] Here, \[{{I}_{t}}=8.96A\] and \[{{I}_{c}}=8A\] \ \[{{\left( \frac{8.96}{8} \right)}^{2}}=1+\frac{{{m}^{2}}}{2}\] or \[1.254=1+\frac{{{m}^{2}}}{2}\] or \[\frac{{{m}^{2}}}{2}=0.254\] or \[{{m}^{2}}=0.508\] or m = 0.71 = 71%
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