Capacitor 1 with a capacitance of 2.00µF is connected in series with capacitor 2

Question

Capacitor 1 with a capacitance of 2.00µF is connected in series with capacitor 2 of capacitance4.00 µF, and a potentialdifference of 200 V is applied acrossthe pair. (a) Calculate the equivalent capacitance.
1 µF

(b) What is the charge on capacitor 1?
2 C

(c) What is the potential difference on capacitor 1?
3 V

(d) What is the charge on capacitor 2?
4 C

(e) What is the potential difference on capacitor 2?
5 V (a) Calculate the equivalent capacitance.
1 µF

(b) What is the charge on capacitor 1?
2 C

(c) What is the potential difference on capacitor 1?
3 V

(d) What is the charge on capacitor 2?
4 C

(e) What is the potential difference on capacitor 2?
5 V

Explanation / Answer

(a)     as the two capacitors are connected in seriesthe equivalent capacitance across them will be     Ceq = (C1 C2) /(C1 + C2) (b)     as the two capacitors are connected in seriesthe charge will be same passing through the both that is theequvalent charge given by     Qeq = Ceq V (c)     the potential difference on capacitor 1      V1 = Qeq /C1 (e)     the potential difference on capacitor 2      V2 = Qeq /C2

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