00:01
In this problem, switch s1 has been closed and s2 has opened for long time and this continues until the capacitors are fully charged and at time t equal to zero, s1 is opened and s2 is closed and we have to find out the current passing through r2 at time t equal to zero in the units of milli -amperes.
00:29
So initially, when the switch s1 is closed, the circuit has r1, c2 and c1.
00:52
So when it is closed for a long time, the two capacitors are fully charged and since they are in series, the charge on them will be equal.
01:03
So, q1 equal to q2 and potential on c1 is q1 upon c1 and potential on c2 is q2 upon c2 and v1 plus v2 will be equal to the potential of the battery because there is zero potential across the resistor when the capacitors are fully charged because at that time, the current is zero in the circuit.
01:37
So v1 plus v2 is q1 upon c1 plus q2 upon c2.
01:45
V1 plus v2 is the emf and q1 is equal to q2.
01:50
So for q2, we can write q1.
01:53
So we have q1, 1 upon c1 plus 1 upon c2 and we put the values.
02:01
Emf is 12v and c1 is 4mf which is 4 times 10 to the power minus 3f and c2 is 8mf which is 8 times 10 to the power minus 3f and from this we get charge q1 equal to 0 .032 coulombs...