00:01
Hello students, here given voltage will be 12 volt, resistor value r1 110 ohm, r2 220 ohm and r3 330 ohm, capacitor value c1 is 50 micro farad and c2 is 40 micro farad.
00:26
Now using kirchhoff law for the first loop we can write v minus i r1 minus i r2 is equal to 0, then v will be equal to i into r1 plus r2, from this current i will be equal to voltage v divided by r1 plus r2.
00:52
Substituting the values here we obtain current i is equal to 0 .036 ampere.
01:00
Now potential difference across the resistor r2 is obtained as v2 is equal to i into r2.
01:08
Substituting for current i and r2 value we obtain voltage across r2 is 8 volt.
01:16
Then after long time capacitors are fully charged and no current will pass across the resistor r3 which is in between c1 and c2.
01:28
Hence voltage across c1 is equal to voltage across c2 minus voltage across r2 that is v2.
01:37
From this we can write qc1 divided by c1 plus qc2 divided by c2 is equal to v2.
01:51
Capacitors are connected in series hence charge on both capacitors are same that is qc1 is equal to qc2.
02:05
Hence taking common and returning qc2 only we have 1 by c1 plus 1 by c2...