Q. The plates of a capacitor are charged to a potential difference of 320 $volts$ and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After $1\ second$ the potential difference between the plates of the capacitor is $240\ volts,$ then after 2 and 3 $seconds$ the potential difference between the plates will be
A) $200$ and $180\ V$
B) $180$ and $135\ V$
C) $160$ and $80\ V$
D) $140$ and $20\ V$
Explanation
(b) $V = V_{0}(e^{- \lambda t})$
After $1\ seconds$
$V_{1} = 320\left( e^{- \lambda} \right) \Rightarrow 240 = 320\left( e^{- \lambda} \right) \Rightarrow e^{- \lambda} = \frac{3}{4}$
After $2\ seconds$
$V_{2} = 320\left( e^{- \lambda} \right)^{2} = 320 \times \left( \frac{3}{4} \right)^{2} = 180\ volt$
After $3\ seconds$
$V_{3} = 320\left( e^{- \lambda} \right)^{3} = 320 \times \left( \frac{3}{4} \right)^{3} = 135\ volt$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.