Question

Two identical capacitors are connected in parallel to an ac generator that has a frequency of 745 hz and produces a voltage of 24 v. the current in the circuit is 0.15

a. what is the capacitance of each capacitor?

Answer 1

In an ac-circuit with capacitors, the Ohm's law can be rewritten as

`V=I X_C`
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:

`X_C = (V)/(I) = (24 V)/(0.15 A)=160 \Omega`

Then, the reactance is related to the equivalent capacitance
`C_(eq)` of the circuit by

`X_C = (1)/(2 \pi f C_(eq))`
Where f=745 Hz is the frequency. Substituting the numbers, we can find
`C_(eq)`:

`C_(eq)= (1)/(2 \pi f X_C)= (1)/(2 \pi (745 Hz)(160 \Omega))=1.34 \cdot 10^(-6)F`

The two capacitors are connected in parallel, so their equivalent capacitance is

`C_(eq)=C_1 + C_2`
But we also know that the capacitors are identical, so C1=C2 (let's call it now C), and therefore

`C_(eq)=2C`
From which we find C, the capacitance of each capacitor:

`C= (C_(eq))/(2) = (1.34\cdot 10^(-6)F)/(2) = 6.7 \cdot 10^(-7)F`