https://r2.erweima.ai/mathbot/image/1725921661857nf8yip6l.jpg

https://r2.erweima.ai/mathbot/image/17259216618583ggmbv4m.jpg

https://r2.erweima.ai/mathbot/image/1725921661858uttr56mk.jpg

por favor puedes hacerlo sin cambiar nada de este problema matemático como tu lo haces

You have uploaded several images. I will solve the problem as it appears in the images without making changes, ensuring consistency with the solution.

### Problem Breakdown:
You have an AC circuit with a resistor, an inductor, and a capacitor connected in series. The given data is:

- Inductance \(L = 2.65 \times 10^{-2}\) H
- Resistance \(R = 10 \, \Omega\)
- Capacitance \(C = 1.00 \times 10^{-2} \, \mu F = 1.00 \times 10^{-8} \, F\)
- Applied Voltage \(V = 100 \, V \, \angle 0^\circ\)
- Frequency \(f = 60 \, Hz\)

### Step-by-Step Calculation:

1. **Angular Frequency (\(\omega\)):**
   The angular frequency \(\omega\) is related to the frequency \(f\) by:
   \[
   \omega = 2 \pi f
   \]
   For \(f = 60 \, Hz\):
   \[
   \omega = 2 \pi \times 60 = 377 \, rad/s
   \]

2. **Inductive Reactance (\(X_L\)):**
   The inductive reactance is given by:
   \[
   X_L = \omega L = 377 \times 2.65 \times 10^{-2} = 9.99 \, \Omega
   \]
   Therefore, the impedance of the inductor is:
   \[
   Z_L = j X_L = j 9.99 \, \Omega
   \]

3. **Capacitive Reactance (\(X_C\)):**
   The capacitive reactance is given by:
   \[
   X_C = \frac{1}{\omega C} = \frac{1}{377 \times 1.00 \times 10^{-8}} = 25.02 \, \Omega
   \]
   Therefore, the impedance of the capacitor is:
   \[
   Z_C = -j X_C = -j 25.02 \, \Omega
   \]

4. **Total Impedance (\(Z_T\)):**
   The total impedance is the sum of the impedances of the inductor, resistor, and capacitor:
   \[
   Z_T = Z_R + Z_L + Z_C = 10 + j 9.99 - j 25.02
   \]
   Simplifying:
   \[
   Z_T = 10 - j 15.03 \, \Omega
   \]

   The magnitude of the total impedance is:
   \[
   |Z_T| = \sqrt{10^2 + (-15.03)^2} = \sqrt{100 + 225.9} = \sqrt{325.9} = 18.05 \, \Omega
   \]
   The phase angle is:
   \[
   \theta_T = \tan^{-1} \left( \frac{-15.03}{10} \right) = -56.31^\circ
   \]
   Therefore, the total impedance is:
   \[
   Z_T = 18.05 \, \Omega \, \angle -56.31^\circ
   \]

5. **Current (\(I\)):**
   The current in the circuit is given by Ohm's law:
   \[
   I = \frac{V}{Z_T} = \frac{100 \, \angle 0^\circ}{18.05 \, \angle -56.31^\circ}
   \]
   Simplifying:
   \[
   I = 5.54 \, A \, \angle 56.31^\circ
   \]

6. **Voltage Across Each Element:**

   - **Voltage across the Inductor (\(V_L\)):**
     \[
     V_L = I Z_L = 5.54 \, A \, \angle 56.31^\circ \times j 9.99 \, \Omega
     \]
     Simplifying the magnitude and phase:
     \[
     V_L = 5.54 \times 9.99 = 55.38 \, V \, \angle (56.31^\circ + 90^\circ) = 55.38 \, V \, \angle 146.31^\circ
     \]

   - **Voltage across the Resistor (\(V_R\)):**
     \[
     V_R = I Z_R = 5.54 \, A \, \angle 56.31^\circ \times 10 \, \Omega
     \]
     Simplifying:
     \[
     V_R = 55.4 \, V \, \angle 56.31^\circ
     \]

   - **Voltage across the Capacitor (\(V_C\)):**
     \[
     V_C = I Z_C = 5.54 \, A \, \angle 56.31^\circ \times (-j 25.02) \, \Omega
     \]
     Simplifying:
     \[
     V_C = 5.54 \times 25.02 = 138.7 \, V \, \angle (56.31^\circ - 90^\circ) = 138.7 \, V \, \angle -33.69^\circ
     \]

### Final Results:

- \(I = 5.54 \, A \, \angle 56.31^\circ\)
- \(V_L = 55.38 \, V \, \angle 146.31^\circ\)
- \(V_R = 55.4 \, V \, \angle 56.31^\circ\)
- \(V_C = 138.7 \, V \, \angle -33.69^\circ\)

These values represent the voltages across the inductor, resistor, and capacitor, respectively.

---

Would you like any further details or explanations on any part of the solution? Here are some related questions that might interest you:

1. How do you calculate the impedance of a series RLC circuit?
2. What happens to the total impedance if the circuit operates at a different frequency?
3. How would the current change if the resistance value increased?
4. Can you explain the significance of phase angles in AC circuits?
5. How does the reactance of inductors and capacitors vary with frequency?

**Tip:** In series RLC circuits, the phase of the current depends on the relative magnitudes of the inductive and capacitive reactances.

En un circuito de corriente alterna se encuentran conectados en serie, una bobina de 2.65x10^-2 H, un condensador cuya capacidad es 1.00x10^-2 µF y una resistencia de 10Ω. Si se conecta el circuito a una línea de diferencia de potencial eficaz igual a 100V, con una frecuencia de 60 ciclos/s. Calcular la diferencia de potencial entre los extremos de cada elemento en el circuito.

Calculate the voltage across each component in a series RLC circuit connected to an AC source with a 100V RMS and 60Hz frequency. The problem involves finding impedance, reactance, and phase angles for inductors, resistors, and capacitors. Suitable for university-level electrical engineering students.

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