To determine the current flow in a circuit that includes a [tex]\(30 \, \mu F\)[/tex] capacitor connected to a [tex]\(240 \, V\)[/tex], [tex]\(60 \, Hz\)[/tex] AC source, we need to follow several steps:
1. Convert Capacitance to Farads:
- The given capacitance is [tex]\(30 \, \mu F\)[/tex] (microfarads).
- We need to convert this to Farads, knowing that [tex]\(1 \, \mu F = 10^{-6} \, F\)[/tex].
- Hence, [tex]\(30 \, \mu F = 30 \times 10^{-6} \, F = 3 \times 10^{-5} \, F\)[/tex].
2. Calculate the Capacitive Reactance ([tex]\(X_C\)[/tex]):
- The formula for the capacitive reactance is given by:
[tex]\[
X_C = \frac{1}{2 \pi f C}
\][/tex]
where:
- [tex]\(X_C\)[/tex] is the capacitive reactance in ohms ([tex]\(\Omega\)[/tex]),
- [tex]\(f\)[/tex] is the frequency in hertz ([tex]\(Hz\)[/tex]),
- [tex]\(C\)[/tex] is the capacitance in farads ([tex]\(F\)[/tex]).
- Substituting the given values [tex]\(f = 60 \, Hz\)[/tex] and [tex]\(C = 3 \times 10^{-5} \, F\)[/tex]:
[tex]\[
X_C = \frac{1}{2 \pi \times 60 \times 3 \times 10^{-5}}
\][/tex]
- This results in [tex]\(X_C \approx 88.42 \, \Omega\)[/tex].
3. Calculate the Current ([tex]\(I\)[/tex]):
- Using Ohm's Law for AC circuits, the current [tex]\(I\)[/tex] through the capacitor is calculated by:
[tex]\[
I = \frac{V}{X_C}
\][/tex]
where:
- [tex]\(I\)[/tex] is the current in amperes (A),
- [tex]\(V\)[/tex] is the voltage in volts (V),
- [tex]\(X_C\)[/tex] is the capacitive reactance in ohms ([tex]\(\Omega\)[/tex]).
- Given [tex]\(V = 240 \, V\)[/tex] and [tex]\(X_C \approx 88.42 \, \Omega\)[/tex]:
[tex]\[
I = \frac{240}{88.42}
\][/tex]
- This results in [tex]\(I \approx 2.714 \, A\)[/tex].
So, the current flow in the circuit is approximately [tex]\(2.714 \, A\)[/tex].
