CBSE Class 12 Physics 2023 Delhi Set 2 Paper

Question : 9

Total: 11

SECTION - C

A current of 1A flows through a coil when it is connected across a DC battery of 100V. If DC battery is replaced by an AC source of 100V and angular frequency 100‌rad‌ s−1, the current reduces to 0.5A. Find
(i) impedance of the circuit.
(ii) self-inductance of coil.
(iii) phase difference between the voltage and the current.

Solution: 👈: Video Solution

(i) Impedance of the circuit =Z=‌

100

0.5

=200Ω
(ii) ‌‌ Resistance ‌‌=R=‌

100

=100Ω
‌‌ Impedance ‌‌=Z=√(ωL)2+R2
‌ Or, ‌‌200‌=√(ωL)2+1002
‌ Or, ‌‌2002‌=(ωL)2+1002
‌ Or, ‌‌√2002−1002‌=ωL
‌ Or, ‌‌173.2‌=100×L
∴‌L‌=1.732H
(iii) ‌‌ Since, ‌‌tan‌θ‌=‌

ωL

‌‌ Or, ‌‌tan‌θ‌=‌

173.2

100

=1.732
‌‌ Or, ‌‌θ‌=tan−11.732
∴‌θ‌=60∘

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