CBSE Class 12 Physics 2023 Delhi Set 2 Paper

Question : 9 of 11

Marks: +1, -0

SECTION - C

A current of

1 {\A}

flows through a coil when it is connected across a DC battery of

100 {\V}

. If DC battery is replaced by an AC source of

100 {\V}

and angular frequency

100 {\rad} {\ s}^{-1}

, the current reduces to

0.5 {\A}

. 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}=200 Ω

(ii)

\;\;\text " Resistance "\;\;=R=\;{100}/{1}=100 Ω

\;\;\text " Impedance "\;\;=Z={√{(ω L)^2+R^2}}

\;\text " Or, "\;\;200\;={√{(ω L)^2+100^2}}

\;\text " Or, "\;\;200^2\;=(ω L)^2+100^2

\;\text " Or, "\;\;{√{200^2-100^2}}\;=ω L

\;\text " Or, "\;\;173.2\;=100 × L

∴ \;L\;=1.732 {\H}

(iii)

\;\;\text " Since, "\;\;\tan θ\;=\;{ω L}/{R}

\;\;\text " Or, "\;\;\tan θ\;=\;{173.2}/{100}=1.732

\;\;\text " Or, "\;\;θ\;=\tan ^{-1} 1.732

∴ \;θ\;=60^{∘}

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