A) 34 mA
B) 31.5 mA
C) 36.5 mA
D) 2.5 mA
Correct Answer: B
Solution :
\[{{i}_{1}}\,=\,\frac{50}{20\times {{10}^{3}}}\,\,\,\Rightarrow \,\,2.5\,mA\] \[i\,\,=\,\,\frac{220-50}{5\times {{10}^{3}}}\,\,\,\Rightarrow \,\,\frac{170}{5\times {{10}^{3}}}\,\,\Rightarrow \,\,32\,mA\] \[{{i}_{1}}+{{i}_{2}}=i\] \[2.5+{{i}_{2}}=32\] \[{{i}_{2}}\,\,=\,\,31.5\,mA\]You need to login to perform this action.
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