Q. At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in second) does the voltage across them become 4 V? (Take ln 5 = 1.6, ln 3 = 1.1)
A) 2
B) 3
C) 2.5
D) $\frac{3}{2}$
Explanation
(a) Voltage across the capacitors will increase from 0 to 10 V exponentially. The voltage at time t will be given by
$V = 10(1 - e^{- t/\tau_{C}})$
Here $\tau_{C} = C_{net}R_{net} = \left( 1 \times 10^{6} \right)\left( 4 \times 10^{- 6} \right) = 4\ s$
$\therefore\ \ V = 10(1 - e^{- t/4})\ $
Substituting $V = 4$ volt, we have,
4=10($1 - e^{- t/4})$
$e^{- t/4} = 0.6 = \frac{3}{5}$
Taking log both sides we have,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \frac{t}{4} = In\ 3 - In\ 5$
or t =$4(In\ 5 - In\ 3) = 2s.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.