Q. 160W-60V lamp is connected at 60 V DC supply. The number of electrons passing through the lamp in 1 min is (the charge of electron $e = 1.6 \times 10^{- 19}C$)
A) $10^{19}$
B) $10^{21}$
C) $1.6 \times 10^{19}$
D) $1.4 \times 10^{20}$
Explanation
(b) Power, $P = \frac{V^{2}}{R}$
$R = \frac{V^{2}}{P} = \frac{(60)^{2}}{160} = 22.5\Omega$
Now, according to Ohm’s law
V=IR
$\therefore\ I = \frac{60}{22.5}$
$\Longrightarrow I = 2.6A$
Here, $t = 60s$
As $I = \frac{ne}{t}$
$\Longrightarrow \ \ n = \frac{I \times t}{e}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{26 \times 60}{1.6 \times 10^{- 19}} \approx 10^{21}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.