Answer:
Option E is correct, 79 μC
Explanation:
Given that,
A capacitor of capacitance C1
C1 = 3 μF
The capacitor is charge to 40V
V1=40V
Another capacitor of capacitance
C2 = 5 μF
Charge to 18V,
V2 =18V
When the positive terminal of one is connected to the negative terminal of the other, then, this is a series connection.
The capacitor are connected in series.
Series connection have the same charge.
Equivalent capacitance for series connection is give as
Ceq = C1 + C2
Ceq = 3μF + 5μF
Ceq = 8μF
Charge in a capacitor is given as
q = CV
Then, charge on Capacitor 1.
q1 = C1V1
q1 = 3μF × 40V
q1 = 120μC
Charge on capacitor 2
q2 = C2V2
q2 = 5μF × 18V
q2 = 90μC
The total charge in the circuit is
Qeq =q1 + q2
Qeq =90 + 120
Qeq = 210 μC
Then, the combine voltage in the circuit is
Qeq = CeqV
Then, V= Qeq/Ceq
V= 210 μC / 8 μC
V=26.25Volts
Then the charge on the 3μF is
q =CV
q =3 μC × 26.25 V
q = 78.75 μC
q ≈ 79 μC
Final answer is E.
