In this circuit, we have a series combination of an inductor and a capacitor, connected to a variable frequency AC source. We are given the following information:

• Inductance $L = 50 \, \text{mH} = 50 \times 10^{-3} \, \text{H}$
• Capacitance $C = 175 \, \mu \text{F} = 175 \times 10^{-6} \, \text{F}$
• Voltage $V = 120 \, \text{V}$
• Current $I = 5 \, \text{A}$

We are tasked with finding the frequency $f$ at which the current is leading (indicative of a capacitive reactance dominating) and has a magnitude of 5 A.

Step 1: Impedance of the RLC circuit

The impedance $Z$ of a series RLC circuit (without resistance) can be expressed as:

$Z = \sqrt{Z_L^2 + Z_C^2}$

where:

• $Z_L = \omega L = 2\pi f L$ is the inductive reactance,
• $Z_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$ is the capacitive reactance.

Step 2: Relating current and voltage

The relationship between the voltage, current, and impedance in an AC circuit is:

$I = \frac{V}{Z}$

Given that $I = 5 \, \text{A}$ and $V = 120 \, \text{V}$, we can solve for the impedance $Z$:

$Z = \frac{V}{I} = \frac{120}{5} = 24 \, \Omega$

Step 3: Solving for the frequency

Now we substitute the expression for the impedance into the formula for $Z$:

$Z = \sqrt{(2\pi f L)^2 + \left( \frac{1}{2\pi f C} \right)^2}$

Substitute the values of $L$ and $C$:

$24 = \sqrt{(2\pi f \times 50 \times 10^{-3})^2 + \left( \frac{1}{2\pi f \times 175 \times 10^{-6}} \right)^2}$

Step 4: Solve the equation

Let's square both sides and simplify:

$576 = (2\pi f \times 50 \times 10^{-3})^2 + \left( \frac{1}{2\pi f \times 175 \times 10^{-6}} \right)^2$

This is a quadratic equation in terms of $f$, and solving it will yield the frequency at which the current is 5 A.

Let me solve that for you. sympy as sp

Define variables

f = sp.symbols('f', real=True, positive=True)

Given values

L = 50e-3 # in Henry C = 175e-6 # in Farads V = 120 # in Volts I = 5 # in Amps

Impedance equation

Z = V / I # in ohms

Inductive reactance and capacitive reactance expressions

Z_L = 2 * sp.pi * f * L Z_C = 1 / (2 * sp.pi * f * C)

Equation for impedance

impedance_eq = sp.Eq(Z, sp.sqrt(Z_L2 + Z_C2))

Solve for frequency f

frequency_solution = sp.solve(impedance_eq, f) frequency_solutionThe two possible frequencies at which the circuit will take a leading current of 5 A are approximately:

1. $f_1 \approx 50.51 \, \text{Hz}$
2. $f_2 \approx 57.31 \, \text{Hz}$

These are the resonant frequencies at which the impedance of the circuit will allow a current of 5 A.

Would you like to explore these solutions further, or do you have any other questions? Here are a few related questions you might find interesting:

1. How does the impedance of the series LC circuit change with frequency?
2. What would happen if we introduced a resistor into the circuit?
3. How does the phase difference between voltage and current change in an LC circuit?
4. How can the resonance frequency be affected by varying the values of $L$ and $C$?
5. What is the effect of a higher capacitance or inductance on the frequency and current?

Tip: For circuits with both inductors and capacitors, the current can either lead or lag the voltage depending on whether the circuit is more inductive or capacitive at the given frequency.