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CBSE Class 12 Physics 2023 Delhi Set 2 Paper

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Question : 9 of 11
Marks: +1, -0
SECTION - C
A current of 1A1 \mathrm{A} flows through a coil when it is connected across a DC battery of 100V100 \mathrm{V}. If DC battery is replaced by an AC source of 100V100 \mathrm{V} and angular frequency 100rads−1100 \mathrm{rad} \mathrm{s}^{-1}, the current reduces to 0.5A0.5 \mathrm{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=1000.5=200Ω=Z=\frac{100}{0.5}=200 \Omega
(ii)  Resistance =R=1001=100Ω\text{ Resistance }=R=\frac{100}{1}=100 \Omega
 Impedance =Z=(ωL)2+R2\text{ Impedance }=Z=\sqrt{(\omega L)^2+R^2}
 Or, 200=(ωL)2+1002\text{ Or, } 200 =\sqrt{(\omega L)^2+100^2}
 Or, 2002=(ωL)2+1002\text{ Or, } 200^2 =(\omega L)^2+100^2
 Or, 2002−1002=ωL\text{ Or, } \sqrt{200^2-100^2} = \omega L
 Or, 173.2=100×L\text{ Or, } 173.2 = 100 \times L
∴L=1.732H\therefore L = 1.732 \mathrm{H}
(iii)  Since, tan⁡θ=ωLR\text{ Since, } \tan \theta = \frac{\omega L}{R}
 Or, tan⁡θ=173.2100=1.732\text{ Or, } \tan \theta = \frac{173.2}{100}=1.732
 Or, θ=tan⁡−11.732\text{ Or, } \theta = \tan^{-1} 1.732
∴θ=60∘\therefore \theta = 60^{\circ}
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