Two capacitors C 1 = 6 µF and C 2 = 24 µF are independently charged to voltages

Question

Two capacitors C1 = 6 µF and C2 = 24 µF are independently charged to voltages V1 = 9 V and V2 = 27 V respectively. (When entering units, use micro for the metric system prefix µ.)

(i) Once equilibrium is reached, what are the magnitudes of the charges on the two capacitors?

Q1:

Q2:

(ii) What is the potential difference across each capacitor?

(b) Instead of connecting the positive plate of C1 to the positive plate of C2 the two capacitors are connected as shown below. Note that the positive plate of one is connected to the negative plate of the other.

(i) Once equilibrium is reached, what are the magnitudes of the charges on the two capacitors now?

Q1:

Q2:

(ii) What is the potential difference across each capacitor now?

Explanation / Answer

Because the capacitors are connected in parallel, they have the same voltage.
In the first case (a), the two charges combine, in the second case (b), they are reversed wrt each other so what is left is the charge difference.
So (a), the total charge is 6*9 + 24*27 = 702uC

The total capacitance when connected together is 6+24 = 30uF
702uC/30uF = 23.4Volts
Q1=23.4*6=140.4uC

Q2 =23.4*24=561.6uC

Note 140.4+561.6=702

Then (b) the total charge is 24*27 - 6*9 = 594uC
594/30 = 19.8Volts

Q1 =19.8*6 = 118.8uC

Q2 =19.8*24 = 475.2uC

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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