A) 100.1 mA
B) 1000.1 mA
C) 10.01 mA
D) 1.01 mA
Correct Answer: A
Solution :
\[{{I}_{G}}\times G=\left( I-{{I}_{G}} \right)\,S\]Þ I = \[\left( 1+\frac{G}{S} \right){{I}_{G}}\] Þ I = 100.1 mA
You need to login to perform this action.
You will be redirected in
3 sec
