00:01
Late time after which voltage drop across the capacitor will be 15 .0 volts.
00:06
So let's do this.
00:09
First, as we know, eds, sorry, emf of the battery equals to the sum of the voltage drops across the resistance, the resistor, and the capacitor.
00:26
Let's simplify this expression.
00:30
So as we know, voltage drop across the resistor x a function of time equals to the current passing through the resistor as a function of time and multiplied by the resistance.
00:44
Therefore, voltage drop across the capacitor as a function of time equals to emf of the battery minus current multiplied by r.
00:57
Here we know that current as a function of time equals to i0, multiplied by x0, multiplied by exponent negative, power by negative t over tau where i0 is emf over r and tau is rc so here is the equation let's write down the final equation so voltage drop across the capacitor as a function of time equals to emf of the battery minus i0 sorry minus emf of the battery divided by r multiplied by r and multiplied by e power by negative t over r c so here we can get rid of the resistance therefore delta v across the capacitor equals to emf multiplied by 1 minus exponent of negative t over rc.
02:37
So now we have to solve this equation.
02:43
Let's simplify it a little bit.
03:15
So now let's take a logarithm of right and left part of the equations...