CBSE Class 12 Physics 2023 Outside Delhi Set 2 Paper

Question : 11

Total: 13

SECTION - C

A series RL circuit with R=10Ω and L=(‌

100

)‌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: 👈: Video Solution

(a) Given, R‌=10Ω
L‌=(‌

100

)mH
V‌=141sin‌(100πt)
‌ Impedance ‌‌=Z=√R2+(ωL)2
Or, ‌Z=√102+(100π×‌

100

×10−3)2
Or, ‌Z=√102+102
∴ ‌Z=10√2Ω
(b) ‌ Phase angle ‌‌=tan−1‌

ωL

Or, Phase angle =tan−1(‌

100π×

100

×10−3

)
Or, ‌‌ Phase angle =tan−11
∴‌‌ Phase angle =45∘
(c) V0=141V
XL=ωL
‌=100π×‌

100

×10−3
‌=10Ω
I0‌=‌

=‌

141

10√2

Voltage drop across inductor =VL=I0XL
‌‌ Or, ‌‌VL=‌

141

10√2

×10
∴‌VL=100V

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