Answer:
V = 26.6 volts
Explanation:
Let Initial Potential Difference be V.
Charge (Q) accumulated in 30 mF Capacitor =
Q = C * V
Q= 30 * V
Now, Common Potential after connecting to uncharged 10mF capacitor in parallel = 20 Volt
Total Charge =Total Capacity * Common Potential
30 * V = ( 30 + 10) * 20
V = 26.6 volts
The unknown potential across the 30-µF capacitor is 6.67 V.
The given parameters;
Since the potential difference between the two capacitors are different, the two capacitors are in series connection.
In series circuit arrangement, the quantity of charge flowing in each capacitor is the same.
[tex]Q_{30\ \mu F} = Q_{10 \ \mu F}[/tex]
[tex]Q_{10 \ \mu F} = CV = 10\times 10^{-6} \times 20 = 0.0002 \ C[/tex]
The potential difference are different and the total potential is given as;
[tex]V_{T} = V_1 + V_2\\\\V_1 = \frac{Q}{C_1} \\\\V_2 = \frac{Q}{C_2} \\\\V_1+V_2 = \frac{Q}{C_1} + \frac{Q}{C_2}\\\\V_1 + V_2 = Q(\frac{1}{C_1} + \frac{1}{C_2} )\\\\V_1 + 20 = Q(\frac{1}{C_1} + \frac{1}{C_2} )\\\\V_1+ 20 = \frac{Q(C_2+ C_1)}{C_1 C_2} \\\\V_1 = \frac{Q(C_2+ C_1)}{C_1 C_2} - 20[/tex]
[tex]V_1 = \frac{0.0002(10\times 10^{-6}\ + \ 30\times 10^{-6})}{(30\times 10^{-6}) (10\times 10^{-6})} - 20\\\\V_1 = \frac{8\times 10^{-9}}{3\times 10^{-10}} - 20\\\\V_1 = 26.67 - 20\\\\V_1 = 6.67 \ V[/tex]
Thus, the unknown potential across the 30-µF capacitor is 6.67 V.
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