The circuit in the image is an AC circuit with a resistor $R_1$ in series with two capacitors $C_1$ and $C_2$ in parallel. The goal is to calculate the current $I_1$, given the supply voltage $U = 6.6 \angle 0^\circ \text{V}$, frequency $f = 384 \text{ Hz}$, and the values for $R_1 = 620 \, \Omega$, $C_1 = 157 \, \text{nF}$, and $C_2 = 0.7 \, \mu\text{F}$.

Steps:

1. Calculate the capacitive reactance of the capacitors $C_1$ and $C_2$. The formula for capacitive reactance $X_C$ is: $X_C = \frac{1}{2 \pi f C}$ where $f$ is the frequency and $C$ is the capacitance.

2. Combine the capacitors $C_1$ and $C_2$ in parallel. The total equivalent capacitance $C_{\text{eq}}$ is given by: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$ Then, find the equivalent capacitive reactance $X_{C_{\text{eq}}}$ using the same formula for $X_C$.

3. Calculate the total impedance of the circuit, combining the resistor $R_1$ and the equivalent capacitive reactance $X_{C_{\text{eq}}}$: $Z_{\text{total}} = R_1 - j X_{C_{\text{eq}}}$ Here, $j$ represents the imaginary unit for the complex impedance due to the reactance.

4. Calculate the current $I_1$ using Ohm's law for AC circuits: $I_1 = \frac{U}{Z_{\text{total}}}$

Let me calculate each step for you.The current $I_1$ in the circuit is approximately:

• 8.4 mA in magnitude, and
• The phase angle is 37.96°.

Would you like further details on any of the calculations, or have any other questions?

Here are 5 related questions to deepen understanding:

1. How does the phase angle between voltage and current affect power in AC circuits?
2. Why is it important to use complex numbers in AC circuit analysis?
3. How do capacitors behave differently from resistors in AC circuits?
4. What would happen to the current if the frequency of the AC source increased?
5. How can the total impedance be minimized in a circuit with capacitors and resistors?

Tip: Always use radians when dealing with angles in trigonometric or exponential functions involving complex numbers!