Physics Alternating Current ETEA 2016
PMDC Verified Question 95 of 127
The potential difference and the current flowing through a component in an A.C. circuit are given by:
\(V = 5 \cos \omega t\text{ volts}\)
\(i = 2 \sin \omega t\text{ amperes}\)
The power dissipated in the instrument is:
A
Zero Watt
B
10 Watt
C
5 Watt
D
2.5 Watt
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: Zero Watt
Concept:

Average real power in an AC circuit depends on the cosine of the phase angle difference \(\phi\) between voltage and current.

Formula / Reaction:

$$P = V_{\text{rms}} I_{\text{rms}} \cos\phi$$

Solution:

  • Using \(\cos(\omega t) = \sin\left(\omega t + \frac{\pi}{2}\right)\), the voltage equation becomes \(V = 5 \sin\left(\omega t + \frac{\pi}{2}\right)\).


  • Current is \(i = 2 \sin(\omega t)\).


  • The phase difference is \(\phi = 90^\circ\) (\(\pi/2\text{ rad}\)).


  • \(P = V_{\text{rms}} I_{\text{rms}} \cos(90^\circ) = 0\text{ W}\).


Why other options are incorrect:

  • Opt_B: \(10\text{ W}\) is the product of peak values \(V_0 I_0\) assuming \(\cos\phi = 1\).


  • Opt_C: \(5\text{ W}\) is \(\frac{1}{2} V_0 I_0\), valid only for in-phase waveforms.


  • Opt_D: \(2.5\text{ W}\) is an incorrect fraction.

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