Answer: The total charge is equal to 19.8meuC
Explanation:
For explanation see the picture attached
After being disconnected and discharged, C1, C2, and C3, connected in series with one another, with the battery, the charge on Capacitor 1 is 19.8 μC.
The capacitance of a capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference. The capacitance of capacitor.
[tex]C=\dfrac{Q}{V}[/tex]
Here, (Q) is the electric charge and (V) is the potential difference.
Three capacitors C1, C2, and C3 and a battery. If only C1 is connected to the battery, the charge on C1 is 31.6 µC.
[tex]C_1=\dfrac{31.6}{V}[/tex] .......1
When the series combination of C2 and C1 is connected across the battery, the charge on C1 is 22.9 µC.
[tex]C_{12}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\\\dfrac{Q_{12}}{V}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\\[/tex]
Put the value of C(1),
[tex]\dfrac{Q_{12}}{V}=\dfrac{V}{31.6}+\dfrac{1}{C_2}\\\dfrac{22.9}{V}=\dfrac{V}{31.6}+\dfrac{1}{C_2}\\[/tex]
On solving it further,
[tex]C_2=\dfrac{83.17}{V}[/tex] ....2
Next, C3, C1, and the battery are connected in series, resulting in a charge on C1 of 26.0 µC.
[tex]C_{13}=\dfrac{1}{C_1}+\dfrac{1}{C_3}\\\dfrac{Q_{13}}{V}=\dfrac{1}{C_1}+\dfrac{1}{C_3}\\[/tex]
Put the value of C(1),
[tex]\dfrac{Q_{13}}{V}=\dfrac{V}{31.6}+\dfrac{1}{C_3}\\\dfrac{26}{V}=\dfrac{V}{31.6}+\dfrac{1}{C_3}\\[/tex]
On solving it further,
[tex]C_3=\dfrac{146.7}{V}[/tex] ....3
The equivalence capacitance can be given as,
[tex]C_{eq}=\dfrac{V}{31.6}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\\C_{eq}=\dfrac{1}{C_1}+\dfrac{V}{83.17}+\dfrac{V}{146.7}[/tex]
Solve it further, we get,
[tex]C_{eq}=\dfrac{19.8}{V}[/tex]
The charge on C1 is,
[tex]Q_1=C_{eq}V\\Q_1=\dfrac{19.8}{V}V\\Q_1=19.8\rm\; \mu C[/tex]
Hence, after being disconnected and discharged, C1, C2, and C3 and connected in series with one another, with the battery, the charge on Capacitor 1 is 19.8 μC.
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