Answer:
6.05×10⁻³ J
Explanation:
Note: Two capacitors connected in series behaves like two resistors connected in parallel.
Using
1/Ct = 1/C1+1/C2
Ct = (C1×C2)/(C1+C2)............................ Equation 1
Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.
Given: C1 = 25 µF, C2 = 50 µF
Substitute into equation 1
Ct = (25×50)/(25+50)
Ct = 1250/75
Ct = 16.67 µF.
Using
Q = CV.................... Equation 2
Where Q = Charge, V = Voltage.
Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F
Substitute into equation 2
Q = 33(16.67×10⁻⁶)
Q = 5.5×10⁻⁴ C.
Since both capacitors are connected in series, the same amount of charge flows through them.
Using,
E = 1/2Q²/C.................. Equation 3
Where E = Energy stored in the 25-µF capacitor
Given: Q =5.5×10⁻⁴ C, C = 25 µF = 25×10⁻⁶ F
Substitute into equation 3
E = 1/2(5.5×10⁻⁴)²/ 25×10⁻⁶
E = 6.05×10⁻³ J.
