A 13µ F capacitor is joined in series with a 35µ F capacitor and a 12 Volts batt

Question

A 13µ F capacitor is joined in series with a 35µ F capacitor and a 12 Volts battery.

A) What is the potential difference across the first capacitor?
B) What is the potential difference across the second capacitor?
C) What is the charge on the first capacitor?
D) What is the charge on the first capacitor?
E) How much energy is stored by the first capacitor?
F) How much energy is stored by the second capacitor?

Explanation / Answer

Q= C*V Q1 = C1*V1 and Q2 = C2*V2 however Q1=Q2 therefore C1*V1= C2*V2 We know V1 + V2 = 12 so solve for V2 and plug into above equation. V2 = 12 - V1 C1* V1 = C2* (12-V1) --> Solve for V1 V1 = 8.75 V (Part A) Find V2 --> V2 = 12 - 8.75 = 3.25 V (Part B) Part C: Q1 = 13e-6 * 8.75 = 1.14e-4 C Part D: Remember Q1 = Q2, therefore Q2 = 1.14e-4 C Part E: E = (1/2)C*V^2 = (1/2)*(13e-6)(8.75)^2 = 4.98e-4 J Part F: Same equation has above using C2 and V2 -> (1/2)(35e-6)(3.25)^2= 1.85e-4 J

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