CBSE Class 12 Physics 2023 Outside Delhi Set 2 Paper

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Question : 11 of 13
 
Marks: +1, -0
SECTION - C

A series RL circuit with R=10Ω and L=(100π)mH is connected to an ac source of voltage V=141 sin (100πt), where V is in volts and t is in seconds. Calculate
(a) impedance of the circuit
(b) phase angle, and
(c) voltage drop across the inductor
Solution:
(a) Given, R=10Ω
L=(100π)mH
V=141sin(100πt)
Impedance =Z=R2+(ωL)2
Or, Z=102+(100π×100π×10−3)2
Or, Z=102+102
∴ Z=102Ω
(b) Phase angle =tan−1ωLR
Or, Phase angle =tan−1(100π×100π×10−310)
Or, Phase angle =tan−11
∴ Phase angle =45∘
(c) V0=141V
XL=ωL
=100π×100π×10−3
=10Ω
I0=V0Z=141102
Voltage drop across inductor =VL=I0XL
Or, VL=141102×10
∴VL=100V
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