Question

two capacitors c1 =2mueF c2=4mueF are connected in series and a potential difference (p.d) of 1200v is applied across it. the potential difference across 2mueF will be

Answer 1

According to question

Capacitance C1 = 2 mueF
                      C2 = 4 mueF

When Connected in series

Total capacitance across Circuit

1/C = 1/C1 + 1/C2
1/C = 1/4 + 1/2
 1/C = 3/4
So   C= 4/3

Thus C = V/Q
          4/3 = 1200 / Q
Thus   Q = 1200 * 3/4
            Q = 900

So potential difference across 2 mueF,    v = C1 * Q          
                                                                 v = 2 * 900
                                                                    = 1800 V