In an ac-circuit with capacitors, the Ohm's law can be rewritten as
[tex]V=I X_C[/tex]
where V is the voltage, I the current and Xc the capacitive reactance.
From this, we can find the value of the reactance for our circuit:
[tex]X_C = \frac{V}{I} = \frac{24 V}{0.15 A}=160 \Omega [/tex]
Then, the reactance is related to the equivalent capacitance [tex]C_{eq}[/tex] of the circuit by
[tex]X_C = \frac{1}{2 \pi f C_{eq}} [/tex]
Where f=745 Hz is the frequency. Substituting the numbers, we can find [tex]C_{eq}[/tex]:
[tex]C_{eq}= \frac{1}{2 \pi f X_C}= \frac{1}{2 \pi (745 Hz)(160 \Omega)}=1.34 \cdot 10^{-6}F [/tex]
The two capacitors are connected in parallel, so their equivalent capacitance is
[tex]C_{eq}=C_1 + C_2[/tex]
But we also know that the capacitors are identical, so C1=C2 (let's call it now C), and therefore
[tex]C_{eq}=2C[/tex]
From which we find C, the capacitance of each capacitor:
[tex]C= \frac{C_{eq}}{2} = \frac{1.34\cdot 10^{-6}F}{2} = 6.7 \cdot 10^{-7}F[/tex]
