00:01
But a of the given problem, first the equivalent capacitor, capacitance, sorry, for a 2, 4 -macro -ferad capacitor connected in a series like this, is given by 4 -macro -farad divided by 2, that is equal to 2 micro -farad.
00:25
This combination is then connected in parallel with the two other, two microferod capacters, one on each side resulting the equivalent capacitance of c is equal to three times of two microferrard, that is six microferrard.
00:43
This is now seen to be there in a series with another combination, which consists of two three microferral capacitors connected in a parallel, which are themselves equivalent to c -dash, that is 2 times of 3 micro ferrard that is equal 6 microferod.
01:08
Thus the equivalent capacitance of the circuit is c equalent is equal to c times c dash divide by c plus c dash.
01:18
Substituting values we get 6 microferod times 6 microferod divided by 6 plus 6.
01:26
This gives us 3 microferod.
01:28
Equal length capacitance.
01:31
For a part b, let v is equal to 20 volts be the potential difference applied by the battery, then we can write q is equal c equivalent times v, that is a 3 microferrath times 20 volts...