00:02
Hi, in the given problem which is as an extension to the problem number 96.
00:09
Here, the capacitor of flash of the camera is taking a time t is equal to 4 .0 into 10 dash minus 3 seconds to get discharge up to 1 % of the maximum charge stored over it means the instantaneous charge stored over the plates of the capacitor after this much time is 1 % of the maximum charge which can be written as 1 by 100 of q0.
00:45
Now we have to find the time constant of such a circuit.
00:50
So using the equation of a discharging capacitor, equation for the 10 -10 is charged stored over it.
01:09
That equation is given as q is equal to q0, e -res to the power minus t by tau.
01:16
Here this q is q0 by 100 is equal to q0 e -d to the power time is given as minus 4 .0 into 10 dash to power minus 3 seconds divided by time constant.
01:32
So canceling this q -0, this expression may be written as e -res -bar minus 4 .0 into 10 dash bar minus 3 divided by tau is equal to 1 by 100 or we can write it like 0 .01.
01:48
So now finally converting this exponential form into logarithmic form we may write it like log of 0 .01 at base e means natural logarithm of 0 .01 is equal to minus 4 .0 in 10 dash to the power minus 3 divided by tau.
02:12
And the value of this natural log of 0 .01 comes out to be minus 4 .605...