A resistor R and an inductor L are connected in series to a source $\text{V}=\text{V}_0\sin\omega\text{t}.$
Find:
Find:
- Peak value of the voltage drops across R and across L,
- Phase difference between the applied voltage and current. Which of them is ahead?
CBSE 55-1-2 PAPER SET 2020
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$\text{V}=\text{v}_0\sin\omega\text{t}$
$\text{X}_\text{L}=\omega\text{L}$
$\text{z}=\sqrt{\text{R}^2+\text{X}_\text{L}^2}=\sqrt{\text{R}^2+(\omega\text{L)}^2}$
Now,
Current $=\frac{\text{V}_0}{\sqrt{\text{R}^2+(\omega\text{L})^2}}$
Voltage drop accurs, reduse,
$=1\text{X}_\text{L}$
$=\frac{(\omega\text{L)}}{\sqrt{\text{X}^2+\omega^2\text{L}^2}}$
And resistance lead to the Inductance.
$\text{X}_\text{L}=\omega\text{L}$
$\text{z}=\sqrt{\text{R}^2+\text{X}_\text{L}^2}=\sqrt{\text{R}^2+(\omega\text{L)}^2}$
Now,
Current $=\frac{\text{V}_0}{\sqrt{\text{R}^2+(\omega\text{L})^2}}$
- Now, voltage drope accurs resestev = R
Voltage drop accurs, reduse,
$=1\text{X}_\text{L}$
$=\frac{(\omega\text{L)}}{\sqrt{\text{X}^2+\omega^2\text{L}^2}}$
- Phase Difference,
And resistance lead to the Inductance.
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