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CBSE Class 12 Physics 2023 Delhi Set 3 Paper

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Question : 13
Total: 14
SECTION - C

An alternating current I=14sin‌(100πt) A passes through a series combination of a resistor of 30 Ω and an inductor of (‌
2
5π
) H. Taking √2=1.4, calculate the
(i) rms value of the voltage drops across the resistor and the inductor, and
(ii) power factor of the circuit.
Solution:     👈: Video Solution
(i) V‌rms ‌ drop across resistor, R
‌=Irms×R
‌=‌
14
√2
×30
‌=294V
Vrms‌ drop across inductor, ‌L
=Irms×ωL
=(14∕√‌2)×(100π)×(2∕5π)
=392V
(ii) Power factor, R∕Z
‌‌
R
Z
=‌
R
√R2+XL2

‌‌
R
Z
=‌
30
√302+(100π×‌
2
5π
)2
‌=‌
30
√302+402

‌=‌
30
50

‌=0.6
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