A 1.41 µF capacitor charged to 51 V and a 2.49 µF capacitor charged to 31 V are connected to each other, with the two positive plates connected and the two negative plates connected. What is the final potential difference across the 2.49 µF capacitor? Answer in units of V.

Question

Physics

Amaris Esparza

27 September, 04:47

A 1.41 µF capacitor charged to 51 V and a 2.49 µF capacitor charged to 31 V are connected to each other, with the two positive plates connected and the two negative plates connected. What is the final potential difference across the 2.49 µF capacitor? Answer in units of V.

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Answers (2)

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Evelin Farley · Answer 1

Given two capacitor of capacitance

C1=1.41 µF, charge to 51V

C2=2.49 µF, charge to 31V

The charge on each capacitor before connection is

Q=CV

Q1=C1V1

Q1=1.41 µF * 51

Q1=71.91 µC

Also, Q2=C2V2

Q2 = 2.49 µF * 32

Q2=77.19 µC

The two capacitor are connected in parallel, so it equivalent capacitor is given as

Ceq=C1+C2

Ceq=1.41 µF + 2.49µF

Ceq = 3.9 µF

The total charge in the circuit is

Qeq=Q1+Q2

Qeq=71.91 µC+77.19 µC

Qeq = 149.1 µC

Then, the total voltage in the circuit

Q=CV

V=Q/C

V=Qeq/Ceq

V = 149.1 µC / 3.9 µF

V=38.23V

Since the capacitors are parallel, then they have the same voltage because parallel connection of capacitor have same voltage.

The voltage on the 2.49µC capacitor is 38.23Volts

Buchanan · Answer 2

38.23Volts

Explanation:

The formula for calculating the charge on a capacitor is expressed as Q = CV where;

Q is the charge on the capacitor

C is the capacitance of the capacitor

V is the voltage across the capacitor.

If 1.41 µF capacitor is charged to 51V, the charge on the capacitor can be calculated using the expression Q = CV where:

C = 1.41 µF, V = 51V

Q = 1.41 * 10^-6 * 51

Q1 = 7.19*10^5coulombs

Similarly, if 2.49 µF capacitor is charged to 31 V, the charge on the capacitor will be:

Q2 = 2.49*10^-6 * 31

Q2 = 7.72*10^-5coulombs.

If the capacitors are connected to each other with the two positive plates connected and the two negative plates connected, this shows they are connected in parallel to each other.

Effective capacitance will be;

Ceff = C1 + C2 (since they are connected in parallel)

Ceff = 1.41µF + 2.49µF

Ceff = 3.90µF

Total charge Qtotal = Q1+Q2

Qtotal = 7.19*10^-5 + 7.72*10^-5

Qtotal = 14.91 * 10^-5coulombs

The total voltage across the connection V = Qtotal/Ceff

V = 14.91*10^-5/3.90*10^-6

V = 38.23Volts.

Since the same voltage flows across the elements in parallel connected circuit, the final potential difference across the 2.49µF will be the same as the total voltage in the circuit which is 38.23Volts.