Question

Two capacitors of capacitance 600 PF and 900PF are connected in series across a 200V supply calculate a) the effective capacitance of the combination b) the total change stored in the system and c) the p. d across each capacitor ​

Answer 1

Given:

Capacitor = c1 = 600 PF

Capacitor = c2 = 900PF

Supply = 200V

To Find:

Pd across capacitors

Solution:

C1 = 600 = 600 × 10-12F

c2 = 900 = 900 × 10-12F

Since capacitors are in series, thus total capacitance -  

1/c = 1/c1 + 1/c2

C = c1c2/c1+c2  

= 600 × 10-12 × 900 × 10-12 / 600 × 10-12 + 900 × 10-12

= 540000 × 10-12/ 1500 × 10-12

= 360 × 10-12F

In series connection, charge on each capacitor is same, thus q = CV

q = 360 × 10-12F  × 200

q = 720 × 10-10F

P.d across 600F = 720 × 10-10/ 600 × 10-12

= 120 V

P.d across 900F = 720 × 10-10/ 900 × 10-12

= 80 V

Answer: The Potential difference is 120V and 80V.

Answer 2

Answer:

effective capacitance of combination=360×10^-12 F

total charge stored in series combination=7.2×10^8 C

the potential difference across each capacitors are

V1=120volt

V2=80volt