Question

An ac voltage $V = 280 \sin(100\pi t)$ volt is connected across a series LCR circuit in which $R = 400 \Omega$, $L = 5/\pi \text{ H}$ and $C = 50/\pi \mu\text{F}$. Taking $\sqrt{2} = 1.4$, calculate rms value of current that flows in the circuit.

Hint:

Always strictly ensure you do not inadvertently mix peak values with RMS values within a single Ohm's Law equation ($V = IZ$); either use peak entirely for both or use RMS entirely for both.
When specific approximations like $\sqrt{2} = 1.4$ or $\pi = 3.14$ are heavily explicitly provided in the question stem, strictly using them is mandatory to exactly match the marking scheme.

Answer 1

Concept:
• In alternating current (AC) circuit analysis, peak values exclusively describe the absolute maximum instantaneous amplitude reached strictly by the oscillating waveform.

• Root Mean Square (RMS) values are fundamentally far more practical, accurately representing the equivalent constant DC value that would unequivocally deliver the exact same average thermal power.

• Ohm's law cleanly applies to AC circuits by meticulously relating peak current strictly to peak voltage, and RMS current strictly to RMS voltage, universally bridged by the total impedance.

Step 1: Extract Parameters and Previous Results
From the rigorously standard mathematical voltage equation $V = 280 \sin(100\pi t)$, we readily identify the absolute peak driving voltage:
\[ V_m = 280 \text{ V} \]
From our meticulous, detailed calculation in the prior sub-question (I), we robustly established the total operational circuit impedance:
\[ Z = 500 \Omega \]

Step 2: Calculate the Peak Current
To systematically find the robust RMS current, we generally must first accurately locate the absolute peak current $I_m$ physically circulating heavily in the closed circuit.
Using Ohm's law rigidly adapted for peak AC variables, we divide the peak voltage by the total opposing impedance:
\[ I_m = \frac{V_m}{Z} \]
Substitute the known exact numerical values safely into the ratio:
\[ I_m = \frac{280}{500} \]
Simplify the numerical fraction completely to acquire a workable decimal:
\[ I_m = \frac{28}{50} = \frac{56}{100} = 0.56 \text{ A} \]
This signifies that at the exact peak of its sinusoidal cycle, the current reaches exactly 0.56 Amperes.

Step 3: Convert to RMS Current
The operating RMS (Root Mean Square) current is tightly and universally tied mathematically to the peak current strictly by the classic root-two divisor for pure sine waves:
\[ I_{rms} = \frac{I_m}{\sqrt{2}} \]
The question explicitly and kindly instructs us to strictly use the specific numerical approximation $\sqrt{2} = 1.4$.
Substitute the calculated peak current and the provided approximation perfectly into the equation:
\[ I_{rms} = \frac{0.56}{1.4} \]
To execute this division smoothly, aggressively multiply both numerator and denominator exactly by 100:
\[ I_{rms} = \frac{56}{140} \]
Simplify the resulting fraction efficiently by recognizing common multiples:
\[ I_{rms} = \frac{4}{10} = 0.4 \text{ A} \]
The effective, power-producing RMS current reliably flowing through this entire setup is exactly 0.4 Amperes.