Answer:
1.394 V
Explanation:
When a capacitor is connected in series, The sum of a capacitance is given as
1/Ct = 1/C1 + 1/C2 + 1/C3 ............................... Equation 1
Where Ct = Value of the combined capacitor, C1 = first capacitance, C2 = Second capacitance, C3 = Third capacitance.
Given: C1 = 4.3 µF, C2 = 12.6 µF, C3 = 31.2 µF
Substitute into equation 1
1/Ct = 1/4.3 + 1/12.6 + 1/31.2
1/Ct = 0.233 + 0.0794 + 0.0321
1/Ct = 0.3445
Ct = 1/0.3445
Ct = 2.9 µF.
But
Q = CtV .................................. Equation 2
Where
Q = Amount of charge, V = voltage, C = total capacitance
Given: V = 15 V, Ct = 2.9 µF
Substitute into equation 1
Q = 15(v) ×2.9(µF)
Q = 43.5 µC
The voltage across the 31.2 µF is
V₃ = Q/31.2 µF......................... equation 3
Where V₃ = The voltage across the 31.2 µF capacitor.
Note: When capacitors are connected in series, the same quantity of charge flows through them.
V₃ = 43.5 (µC)/31.2 (µF)
V₃ = 1.394 V.
Hence the voltage across the 31.2 µF = 1.394 V
