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 : 5=1.6, 3=1.1]

Step-by-step solution

Voltage across the capacitors will increase from 0 to 10 ~V exponentially. The voltage at time t will be given by aligned & V=10 (1-e^ -t / C ) & Here _c=C_ net R_ net & = (1 10^6 ) (4 10⁻⁶ )=4 ~s & V=10 (1-e^ -t / 4 ) & aligned Substituting V=4 volt we have, 4=10 (1-e^ -t / 4 ) Substituting V=4 volt we have, 4=10 (1-e^ -t / 4 ) or e^ -t / 4 =0.6= 3 5 Taking both sides we have, - t 4 = 3- 5 or t=4( 5- 3)=2 ~s . Hence, the answer is 2 .