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Exam
NEET
Subject
Physics
Chapter
Electromagnetic Induction

NEET · Physics · Electromagnetic Induction

An AC voltage source given by v = 10 (ω t) V is connected in series...

An AC voltage source given by v=10sin⁡(ωt) Vv = 10 \sin(\omega t)\text{ V} is connected in series with a DC battery of 15 V15\text{ V}. What is the nature of the resultant voltage across a load resistor connected to this combination?

Options

  1. AAlternating voltage
  2. BSteady DC voltage
  3. CSymmetric DC voltage
  4. DPulsating DC voltage

Correct answer

D. Pulsating DC voltage

Step-by-step solution

The resultant voltage across the load is the sum of the DC and AC components: V(t)=15+10sin⁡(ωt)V(t) = 15 + 10 \sin(\omega t).

The maximum value of this voltage is 15+10=25 V15 + 10 = 25\text{ V} and the minimum value is 15−10=5 V15 - 10 = 5\text{ V}.

Since the minimum voltage is positive (+5 V+5\text{ V}), the resultant voltage never crosses zero and never reverses its polarity. An alternating voltage must periodically reverse its direction. Because this voltage remains unidirectional but varies in magnitude, it is classified as a pulsating DC voltage.

Answer: Pulsating DC voltage
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