Solution for Part (a)
Step 1 (Governing Laws & Phasor Setup): [1 Mark]
For a series LCR circuit, the applied voltage $V = V_m \sin \omega t$ is the phasor sum of voltages across the resistor ($V_R$), inductor ($V_L$), and capacitor ($V_C$).
- $V_R = IR$ (in phase with $I$)
- $V_L = IX_L$ (leads $I$ by $\pi/2$)
- $V_C = IX_C$ (lags $I$ by $\pi/2$)
Step 2 (Mathematical Derivation of Impedance): [1.5 Marks]
From the phasor diagram, the resultant voltage $V$ is given by:
$$V^2 = V_R^2 + (V_L - V_C)^2$$Substituting $V = IZ$, $V_R = IR$, $V_L = IX_L$, and $V_C = IX_C$:
$$(IZ)^2 = (IR)^2 + (IX_L - IX_C)^2$$ $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$Where $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$.
Step 3 (Current Expression & Phase Relationship): [1 Mark]
The instantaneous current is $I = I_m \sin(\omega t - \phi)$, where $I_m = \frac{V_m}{Z}$.
The phase angle $\phi$ is given by:
$$\tan \phi = \frac{X_L - X_C}{R}$$If $X_L > X_C$, current lags voltage. If $X_C > X_L$, current leads voltage.
Step 4 (Power Factor Definition & Conditions): [1.5 Marks]
Definition: Power factor ($\cos \phi$) is the ratio of resistance to impedance ($R/Z$).
- Maximum: $\cos \phi = 1$ when $\phi = 0$ (Resonance condition: $X_L = X_C$).
- Minimum: $\cos \phi = 0$ when $\phi = \pi/2$ (Purely inductive or capacitive circuit).
💡 EL10 Concept Insight
Think of the LCR circuit like a tug-of-war: the inductor pulls the current one way, the capacitor pulls it the other, and the resistor acts like friction slowing everything down; the "impedance" is just the total combined resistance to the flow of electricity.
Solution for Part (b)
Step 1 (Inductor Phase Proof): [2 Marks]
For an ideal inductor, $V = L \frac{di}{dt}$. If $i = I_m \sin \omega t$, then:
$$V = L \frac{d}{dt} (I_m \sin \omega t) = \omega L I_m \cos \omega t$$ $$V = V_m \sin(\omega t + \pi/2)$$Since the argument is $(\omega t + \pi/2)$, the voltage leads the current by $\pi/2$.
Step 2 (Induced EMF Graph): [1.5 Marks]