Answer:
[tex]V_G=3.302857143V[/tex]
Explanation:
Let´s find the capacitor impedance:
[tex]V=IZ_C[/tex]
Using the data provided:
[tex]Z_C=\frac{V}{I}=\frac{2}{0.035}=57.14285714[/tex] Ω
Now let´s find the capacitance:
[tex]Z_C=\frac{1}{j2\pi fC }[/tex]
[tex]C=\frac{1}{j2\pi fZ_C} =\frac{1}{j2\pi (3400)(57.14285714)}= 0.8191798542[/tex] μF
We can calculate the capacitor impedance for the second case:
[tex]Z_C=\frac{1}{j2\pi fC}=\frac{1}{j2\pi (5000)(0.8191798542X10^{-6}) }=38.85714285[/tex] Ω
Finally we can calculate the voltage of the second generator:
[tex]V_G=IZ_C=(0.085)(38.85714285)=3.302857143V[/tex]