Question

For the circuit shown below, instantaneous current through inductor '$L$' and capacitor '$C$' is respectively

• A. $-\frac{e_0}{\omega L} \cos \omega t \ ; \ e_0 \omega C \cos \omega t$
• B. $-\frac{e_0}{\omega L} \sin \omega t \ ; \ e_0 \omega C \cos \omega t$
• C. $\frac{e_0}{\omega L} \cos \omega t \ ; \ \frac{e_0}{\omega C} \sin \omega t$
• D. $-\frac{e_0}{\omega L} \sin \omega t \ ; \ \frac{e_0}{\omega C} \cos \omega t$

Hint:

Remember the mnemonic ELI the ICE man: In an inductor (L), Voltage (E) leads Current (I). In a capacitor (C), Current (I) leads Voltage (E). Since the input voltage is a sine function, the lagging current must be $-\cos \omega t$ and the leading current must be $+\cos \omega t$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the instantaneous current functions flowing through an ideal inductor $L$ and an ideal capacitor $C$ when connected independently across an alternating electromotive voltage source given by $e = e_0 \sin \omega t$.

Step 2: Key Formula or Approach:
We apply the standard phase relationship rules for purely reactive AC circuit elements: In a purely inductive circuit, the alternating current lags behind the applied voltage phase by exactly $90^\circ$ ($\frac{\pi}{2}$ radians). In a purely capacitive circuit, the alternating current leads the applied voltage phase by exactly $90^\circ$ ($\frac{\pi}{2}$ radians). The peak current values are determined by the reactances: $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$.

Step 3: Detailed Explanation:
Let the alternating voltage source be $e = e_0 \sin \omega t$. 1.

For the Inductor ($L$): The alternating current lags by $\frac{\pi}{2}$ radians: $$i_L = \frac{e_0}{X_L} \sin\left(\omega t - \frac{\pi}{2}\right) = \frac{e_0}{\omega L} (-\cos \omega t) = -\frac{e_0}{\omega L} \cos \omega t$$ 2.

For the Capacitor ($C$): The alternating current leads by $\frac{\pi}{2}$ radians: $$i_C = \frac{e_0}{X_C} \sin\left(\omega t + \frac{\pi}{2}\right) = \frac{e_0}{\left(\frac{1}{\omega C}\right)} \cos \omega t = e_0 \omega C \cos \omega t$$ Combining these two derived expressions gives the pair: $-\frac{e_0}{\omega L} \cos \omega t$ and $e_0 \omega C \cos \omega t$. This matches option (A).

Step 4: Final Answer:
The currents are $-\frac{e_0}{\omega L} \cos \omega t$ and $e_0 \omega C \cos \omega t$ respectively, which corresponds to option (A).