A) 490 W
B) 790 W
C) 590 W
D) 990 W
Correct Answer: B
Solution :
\[E=\frac{e}{(R+{{R}_{h}}+r)}.\frac{R}{L}\times l\] \[\Rightarrow 10\times {{10}^{-3}}=\frac{2}{(10+R+0)}\times \frac{10}{1}\times 0.4\]Þ R = 790WYou need to login to perform this action.
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