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
The voltage across the 31.2-µF capacitor will be 1.394 V
For the capacitor connected in series
[tex]\dfrac{1}{C_t} =\dfrac{1}{C_1} +\dfrac{1}{C_2} +\dfrac{1}{C_3}[/tex] .............................(1)
Where
[tex]C_t[/tex]= Value of the combined capacitor,
[tex]C_1[/tex]= first capacitance,
[tex]C_2[/tex] = Second capacitance,
[tex]C_3[/tex] = Third capacitance.
Given: C1 = 4.3 µF, C2 = 12.6 µF, C3 = 31.2 µF
Substitute into equation 1
[tex]\dfrac{1}{C_t} =\dfrac{1}{ 4.3} +\dfrac{1}{12.6} +\dfrac{1}{31.2}[/tex]
[tex]\dfrac{1}{C_t} =0.233+0.0794+0.0321[/tex]
[tex]\dfrac{1}{C_t} =0.3445[/tex]
[tex]C_t=\dfrac{1}{0.3445}[/tex]
[tex]C_t=2.9 \mu F[/tex]
But
[tex]Q= C_tV[/tex]Q .................................. (2)
Where
Q = Amount of charge,
V = voltage,
C = total capacitance
Given: V = 15 V, Ct = 2.9 µF
Substitute into equation 1
[tex]Q=15(V)\times 2.9\mu F[/tex]
[tex]Q= 43.5\mu F[/tex]
The voltage across the 31.2 µF is
[tex]V_3=\dfrac{Q}{31.2} \mu F[/tex]
Where V₃ = The voltage across the 31.2 µF capacitors.
Note: When capacitors are connected in series, the same quantity of charge flows through them.
[tex]V_3=\dfrac{43.5}{31.2}[/tex]
[tex]V_3=1.394V[/tex]
The voltage across the 31.2-µF capacitor will be 1.394 V
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