A) \[{{10}^{19}}\]
B) \[{{10}^{21}}\]
C) \[1.6\times {{10}^{19}}\]
D) \[1.4\times {{10}^{20}}\] (The charge of electron \[1.6\times {{10}^{-19}}C)\]
Correct Answer: B
Solution :
Power, \[P=\frac{{{V}^{2}}}{R}\] \[R=\frac{{{V}^{2}}}{P}\] \[=\frac{{{(60)}^{2}}}{160}=22.5\Omega \] 'view. according to Ohm's law \[V=IR\] \[\therefore \] \[I=\frac{60}{22.5}\] \[\Rightarrow \] \[I=2.6A\] Here, \[t=60s\] As \[I=\frac{ne}{t}\] \[\Rightarrow \] \[n=\frac{I\times t}{e}\] \[=\frac{2.6\times 60}{1.6\times {{10}^{-19}}}\] \[\approx {{10}^{21}}\]
You need to login to perform this action.
You will be redirected in
3 sec
