A) 1.6 amp to the left
B) 1.6 amp to the right
C) 0.45 amp to the right
D) 0.45 amp to the left
Correct Answer: B
Solution :
Net current \[{{i}_{net}}={{i}_{(+)}}+{{i}_{(-)}}\] \[=\frac{{{n}_{(+)}}{{q}_{(+)}}}{t}+\frac{{{n}_{(-)}}{{q}_{(-)}}}{t}\] \[=\frac{{{n}_{(+)}}}{t}\times 2e+\frac{{{n}_{(-)}}}{t}\times e\] \[=\text{ }3.2\times {{10}^{18}}\times 2\times 1.6\times {{10}^{19}}+\text{ }3.6\times {{10}^{18}}\times 1.6\times {{10}^{19}}\] = 1.6 A (towards right)You need to login to perform this action.
You will be redirected in
3 sec