If you have five capacitors with capacitances 1.1×10-6 F, 2.3×10-6 F, 3.9×10-6 F

Question

If you have five capacitors with capacitances 1.1×10-6 F, 2.3×10-6 F, 3.9×10-6 F, and two 9.5×10-6 F in series.

1) What is the equivalent capacitance of all five? C = F

2) Initially the capacitors are uncharged. Now a 11 V battery is attached to the system. How much charge is on the positive plate of the 3.9×10-6 F capacitor? Q = C

3) What is the potential difference between the plates of the 3.9×10-6 F capacitor? V = V

4) How much energy is stored in the entire capacitor system? PE = J

5) If you have five capacitors with capacitances 1.1×10-6 F, 2.3×10-6 F, 3.9×10-6 F, and two 9.5×10-6 F in parallel. What is the equivalent capacitance of all five? What is the equivalent capacitance of all five? C = F

6) If one attaches a 11 V battery is attached to the system. How much charge is on the positive plate of the 3.9×10-6 F capacitor? Q = C

7) What is the potential difference between the plates of the 9.5×10-6 F capacitor? V = V

8) How much energy is stored in the entire capacitor system? PE = J

Explanation / Answer

1) In series 1/Ceq = 1/1.1 × 10-6 F + 1/2.3 × 10-6 F + 1/3.9 × 10-6 F + 1/9.5 × 10-6 F + 1/9.5 × 10-6 F..

So 1/Ceq = 1810810.09

So Ceq = 0.552 x10^-6 F

2)In series the charge is the same on each capacitor...so using q = CV

we have q = 0.552x10^-6*11 = 6.07 x10^-6 C

3) V = q/C = 6.07x10^-6/3.9x10^-6 = 1.56 V

4) E = 1/2*Ceq*V^2 = 1/2*0.552x10^-6*11^2 = 3.34x10^-5 J


5) Now in parallel the Ceq = sum of the capacitor =
1.1 × 10-6 F + 2.3 × 10-6 F + 3.9 × 10-6 F + 9.5 × 10-6 F + 9.5 × 10-6 F = 26.3x10^-6F

6) Now the voltage is the same over each so q = C*V = 3.9x10^-6*11 = 4.29x10^-5 C

7)  V = q/C = 4.29x10^-5/9.5x10^-6 = 4.52 V

8) E = 1/2*Ceq*V^2 = 0.5*26.3x10^-6*11^2 = 1.59x10^-3 J

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