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Alternating Current — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current3 Marks
Question
A resistor R and an inductor L are connected in series to a source $\text{V}=\text{V}_0\sin\omega\text{t}.$
Find:
  1. Peak value of the voltage drops across R and across L,
  2. Phase difference between the applied voltage and current. Which of them is ahead?
✓

Answer

$\text{V}=\text{v}_0\sin\omega\text{t}$
$\text{X}_\text{L}=\omega\text{L}$
$\text{z}=\sqrt{\text{R}^2+\text{X}_\text{L}^2}=\sqrt{\text{R}^2+(\omega\text{L)}^2}$
Now,
Current $=\frac{\text{V}_0}{\sqrt{\text{R}^2+(\omega\text{L})^2}}$
  1. Now, voltage drope accurs resestev = R
$=\frac{\text{V}_0\text{R}}{\sqrt{\text{R}^2+\omega^2\text{L}^2}}$

Voltage drop accurs, reduse,

$=1\text{X}_\text{L}$

$=\frac{(\omega\text{L)}}{\sqrt{\text{X}^2+\omega^2\text{L}^2}}$
  1. Phase Difference,
$\tan\theta=\frac{\text{X}_\text{L}}{\text{R}}$

And resistance lead to the Inductance.

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