• # question_answer 160W-60V lamp is connected at 60 V DC Supply. The number of electrons passing trough the lamp in 1 min is A) ${{10}^{19}}$                                   B) ${{10}^{21}}$ C) $1.6\times {{10}^{19}}$                              D) $1.4\times {{10}^{20}}$ (The charge of electron $1.6\times {{10}^{-19}}C)$

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           $I=\frac{ne}{t}$ $\Rightarrow$               $n=\frac{I\times t}{e}$                 $=\frac{2.6\times 60}{1.6\times {{10}^{-19}}}$                 $\approx {{10}^{21}}$