Answer:
Explanation:
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
Answer:
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.
