Complete Question
The complete question is shown on the first uploaded image
Answer:
The final voltage across X is [tex]V_X = 94 V[/tex]
Explanation:
First we are going to obtain the equivalent capacitance for the whole circuit
to do this we will first find the capacitance of the capacitors connected in series
This can be mathematically evaluated as
[tex]\frac{1}{C_s} = \frac{1}{C_Y} + \frac{1}{C_Z}[/tex]
Substituting values
[tex]C_s = \frac{3 \mu F * 3 \mu F }{(3 \mu F ) + (3 \mu F)}[/tex]
[tex]C_s = \frac{9 \mu F ^2}{6 \mu F}[/tex]
[tex]C_s =1.5 \mu F[/tex]
Now the equivalent capacitance of the capacitors connected in series [tex]C_s[/tex] which is connected in parallel to [tex]C_X[/tex]
So the equivalent capacitance of the whole circuit is mathematically evaluated as
[tex]C_T = C_X + C_s[/tex]
Substituting values
[tex]C_T = 1.5 + 4[/tex]
[tex]C_T = 5.5 \mu F[/tex]
Now the total charge stores in these capacitors is mathematically evaluated as
[tex]Q = C_{T} V_{ab}[/tex]
Substituting values
[tex]Q = 5.5 \mu * 120[/tex]
[tex]Q = 660 \mu C[/tex]
When Switch [tex]S_2[/tex] is closed the capacitor [tex]C_Y[/tex] do not current anymore which implies that it has been excluded from the circuit
Hence the equivalent capacitance becomes
[tex]C_T__{1}} = C_Z + C_X[/tex]
[tex]C_T__{1}} = 3 \mu F + 4 \mu F[/tex]
[tex]C_T__{1}} =7 \mu F[/tex]
Now at the point [tex]S_2[/tex] is closed the Voltage across the capacitor [tex]C_X[/tex] sis mathematically represented as
[tex]V_X = \frac{Q}{C_T__{1}}}[/tex]
[tex]V_X = \frac{600 \mu C}{7 \mu F}[/tex]
[tex]V_X = 94.2 V[/tex]
[tex]V_X = 94 V[/tex]
