Question

(i) Define the terms (I) resonant frequency, and (II) power factor of a series LCR circuit. For what value of the power factor will the power dissipated in the circuit be maximum?

(ii) An inductor of $\dfrac{5}{\pi}$ H, a capacitor of $\dfrac{50}{\pi}\ \mu\text{F}$ and a resistor of $400\ \Omega$ are connected in series across an ac voltage $v = 140 \sin(100\pi t)$ V. Calculate:

(I) impedance of the circuit, and

(II) rms value of current that flows in the circuit.

(Take $\sqrt{2} = 1.4$)

Hint:

Define the two terms, then use X_L = wL, X_C = 1/(wC), Z = sqrt(R^2 + (X_L - X_C)^2) and I_rms = V_rms/Z with V_rms = 140/1.4.

Answer 1

Step 1: What is asked:
We must define resonant frequency and power factor, say when power is largest, and then find the impedance and the rms current for the given series LCR circuit.

Step 2: Resonant frequency in words:
A coil opposes a changing current with reactance $\omega L$. A capacitor opposes it with reactance $1/\omega C$. The two act in opposite senses.
At the frequency where they are equal, they cancel. The circuit then offers only the resistance $R$, so the current peaks.
That frequency is $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$ and it is called the resonant frequency.

Step 3: Power factor in words:
Current lags or leads the voltage by an angle $\phi$. The factor $\cos\phi = R/Z$ tells what fraction of the apparent power $V_{rms}I_{rms}$ is really used up as heat in the resistor.

Step 4: Maximum power:
Since $P = V_{rms}I_{rms}\cos\phi$ and $\cos\phi$ can be at most 1, the power is maximum for power factor $= 1$. Then $\phi = 0$ and $Z = R$.

Step 5: Work with the peak values first:
Here $\omega = 100\pi$ rad/s and $V_0 = 140$ V. The $\pi$ in the data cancels the $\pi$ in $\omega$.
$X_L = 100\pi \times \dfrac{5}{\pi} = 500\ \Omega$ and $X_C = \dfrac{1}{100\pi \times (50/\pi)\times 10^{-6}} = 200\ \Omega$, so the net reactance is $X = 300\ \Omega$.
The reactance is inductive, because $X_L > X_C$.

Step 6: Impedance and peak current:
The sides $400$ and $300$ form a 3-4-5 triangle scaled by 100, so
\[ Z = \sqrt{400^2 + 300^2} = 500\ \Omega \]
\[ I_0 = \frac{V_0}{Z} = \frac{140}{500} = 0.28\ \text{A} \]

Step 7: Convert to rms:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{0.28}{1.4} = 0.2\ \text{A} \]
This agrees with $V_{rms}/Z = 100/500$, which is a good check.

Final Answer:
Resonance occurs where the reactances cancel, power factor is R/Z, and power peaks at power factor 1. The impedance is 500 ohm and the rms current is 0.2 A. \[ \boxed{Z = 500\ \Omega,\quad I_{rms} = 0.2\ \text{A}} \]