To solve the problem, we need to analyze the given series RLC circuit with the following parameters:
- Frequency, \( f = 1000 \, \text{Hz} \)
- Inductance, \( L = 90.0 \, \text{mH} = 90 \times 10^{-3} \, \text{H} \)
- Capacitance, \( C = 0.5 \, \mu\text{F} = 0.5 \times 10^{-6} \, \text{F} \)
- Phase constant, \( \phi = 60^\circ \)
We need to determine whether the current leads the voltage or vice versa, and also find the resistance and the resonant frequency.
### Step 1: Calculate the inductive reactance (\( X_L \))
The inductive reactance is given by the formula:
\[
X_L = 2 \pi f L
\]
Substituting the values:
\[
X_L = 2 \pi (1000) (90 \times 10^{-3}) = 180 \pi \, \Omega
\]
### Step 2: Calculate the capacitive reactance (\( X_C \))
The capacitive reactance is given by the formula:
\[
X_C = \frac{1}{2 \pi f C}
\]
Substituting the values:
\[
X_C = \frac{1}{2 \pi (1000) (0.5 \times 10^{-6})} = \frac{1}{\pi \times 10^{-3}} = \frac{1000}{\pi} \, \Omega
\]
### Step 3: Compare \( X_L \) and \( X_C \)
Now we compare \( X_L \) and \( X_C \):
- \( X_L = 180 \pi \approx 565.2 \, \Omega \)
- \( X_C = \frac{1000}{\pi} \approx 318.47 \, \Omega \)
Since \( X_L > X_C \), the circuit is inductive, which means that the voltage leads the current.
### Step 4: Calculate the resistance (\( R \))
Using the phase angle relation:
\[
\tan(\phi) = \frac{X_L - X_C}{R}
\]
We know \( \phi = 60^\circ \), thus:
\[
\tan(60^\circ) = \sqrt{3}
\]
Substituting the values:
\[
\sqrt{3} = \frac{180 \pi - \frac{1000}{\pi}}{R}
\]
Rearranging gives:
\[
R = \frac{180 \pi - \frac{1000}{\pi}}{\sqrt{3}}
\]
Substituting \( \pi^2 = 10 \):
\[
R = \frac{1800 - 1000}{\sqrt{3} \cdot \pi} = \frac{800}{\sqrt{3} \cdot \pi}
\]
This simplifies to:
\[
R = \frac{80 \pi}{\sqrt{3}} \, \Omega
\]
### Step 5: Calculate the resonant frequency (\( \omega \))
At resonance, the formula for angular frequency is:
\[
\omega = \frac{1}{\sqrt{LC}}
\]
Substituting the values:
\[
\omega = \frac{1}{\sqrt{(90 \times 10^{-3})(0.5 \times 10^{-6})}} = \frac{1}{\sqrt{45 \times 10^{-9}}} = \frac{10^4}{3} \sqrt{2}
\]
Thus, we have:
\[
\omega = \frac{\sqrt{2}}{3} \times 10^4 \, \text{rad/s}
\]
### Summary of Results
- **Current leads the voltage**: **No**, **Voltage leads the current**.
- **Resistance \( R \)**: \( \frac{80 \pi}{\sqrt{3}} \, \Omega \).
- **Resonant frequency \( \omega \)**: \( \frac{\sqrt{2}}{3} \times 10^4 \, \text{rad/s} \).
