A) \[1.25\times {{10}^{6}}\]
B) \[\frac{G}{19}\]
C) \[\frac{G}{20}\]
D) \[E=16T-0.04{{T}^{2}}\mu V.\]
Correct Answer: A
Solution :
Here, \[\left( \frac{0.4}{\pi } \right)\] ......(i) \[11.4\Omega ,15.5A\] and image is real, magnification is negative \[30.7\Omega ,6.5A\] From, - \[40.4\Omega ,5A\] \[50\Omega ,5A\] \[{{R}_{1}}\] Put in Eq. (i) \[{{R}_{2}}\] \[{{R}_{1}}>{{R}_{2}}\]
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