00:01
Why, given the voltage versus time of a capacitor during its charging and it is charging through a resistor r is equal to 4 ,700 om.
00:18
In the first part we are asked to calculate the emf of the battery.
00:25
From the graph we can see that the maximum voltage attained by the capacitor is the steady value which is 9 volt and this steady value is the supply of.
00:36
Voltage so we can say voltage of the battery e is equal to 9 volt.
00:45
In the second part we are asked to calculate the time constant to.
00:50
Time constant is the time taken for a capacitor to charge 63 percentage of the maximum value.
00:59
So let us note the time for voltage reaching its 63 percentage of the maximum value.
01:05
So we know the maximum value vm is equal to 9.
01:09
So 63 percentage of vm will be equal to 0 .63 into 9 volt which is equal to 5 .67 volt.
01:22
From the graph the time corresponding to the voltage of 5 .67 that is this point will be approximately 1 .5 milliseconds.
01:35
Hence we can say that the time constant toe is equal to 1 .5 millie seconds.
01:41
The charging of a capacitor can be expressed using the formula v is equal to emfe into 1 minus e raise to minus t by rc...