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

The equivalent capacitance ceq = c1*c2 / (c1 + c2)= 2.0F * 4.0F / (2.0F + 4.0F) =1.33F (b) The charge on capacitor   c1            q =ceq V = 1.33F * 200V = 266 C (c) The potential difference V1 = q/c1 = 266C / 2F =133 V (d) The charge on capacitor c2 q = 266C (e) The potential difference V2 = q/c2 = 66.5 V

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.