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.
