Physics Alternating Current PMDC Conceptual Practice
PMDC Verified Question 49 of 127
A \(100\ \Omega\) pure resistor is connected across an AC voltage source \(V = 200 \sin(100\pi t)\text{ V}\). What is the peak current \(I_0\) flowing through the resistor?
A
2 A
B
\(1.414\text{ A}\)
C
4 A
D
200 A
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: 2 A
Concept:

In a pure resistor, Ohm's law applies directly to peak values: \(I_0 = \frac{V_0}{R}\).

Formula / Reaction:

$$I_0 = \frac{V_0}{R}$$

Solution:

  • From the equation: \(V_0 = 200\text{ V}\).


  • Resistance: \(R = 100\ \Omega\).


  • \(I_0 = \frac{200\text{ V}}{100\ \Omega} = 2\text{ A}\).


Why other options are incorrect:

  • Opt_B: \(1.414\text{ A}\) is the RMS current \(I_{\text{rms}} = \frac{2}{\sqrt{2}}\text{ A}\).


  • Opt_C: \(4\text{ A}\) results from using half resistance.


  • Opt_D: \(200\text{ A}\) represents voltage without division.

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