Answer:
The value is [tex] C_u = 5 *10^{-6} F[/tex]
Explanation:
From the question we are told that
The voltage of the capacitor is [tex]V = 90 \ V[/tex]
The capacitance of the capacitor is [tex]C = 30 \ \mu F = 30 *10^{-6} \ F[/tex]
The final voltage is [tex]V_a = 20 \ V[/tex]
Since the capacitors are connected in parallel we have that
[tex]Q_u = Q'_u + Q[/tex]
Where [tex]Q_u[/tex] is the charge of the known capacitor before it is connected to the known capacitor
[tex]Q_u' [/tex] is the charge of the known capacitor after it is connected
[tex]Q [/tex] is the charge of the unknown capacitor
Also the potential across the two capacitors will be the same (except for a loss due heat (it been converted to heat ))
So
[tex]CV = CV_a + C_u * V_a[/tex]
=> [tex] C_u * V_a = CV - CV_a [/tex]
=> [tex] C_u = \frac{CV - CV_a }{ V_a } [/tex]
=> [tex] C_u = \frac{ [30*10^{-6} *90] - [30*10^{-6} * 20] }{ 20 } [/tex]
=> [tex] C_u = 5 *10^{-6} F[/tex]
