MCQ
The average value of potential difference $V$ shown in figure is
- A$V_0$
- B$\frac{V_0}{2}$
- ✓$\frac{V_0}{4}$
- D$\frac{2V_0}{3}$
✓
Answer
Correct option: C.
$\frac{V_0}{4}$
c
$<\mathrm{V}>=\frac{\text { Area of graph }}{\text { Time }} $
$<\mathrm{V}>=\frac{\text { Area of graph }}{\text { Time }} $
$=\frac{\frac{1}{2} \times \frac{\mathrm{T}}{2} \times \mathrm{V}_{0}}{\mathrm{T}}=\frac{\mathrm{V}_{0}}{4} $
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