Question) An LR circuit contains a resistance of 40 mu and an inductance of 25 mH. What would be the power dissipated in the circuit for an applied alternating voltage of frequency 1000 Hz and peak value 10 V?

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Solution

Dear student
$a s s u m i n g 40 Ω r e s i s \tan c e Z = \sqrt{{X_{L}}^{2} + R^{2}} = \sqrt{ω^{2} L^{2} + R^{2}} = \sqrt{4 π^{2} f^{2} L^{2} + R^{2}} = 6280.12 Ω \cos θ = \frac{R}{Z} = \frac{40}{6280.12} = 0.0063 I_{0} = \frac{V}{Z} = \frac{10}{6280.12} = 0.0015 P = \frac{V_{0} I_{0} \cos θ}{2} = \frac{10 \times 0.0015 \times 0.0063}{2} = 0.0000047 W R e g a r d s$

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Question) An LR circuit contains a resistance of 40 mu and an inductance of 25 mH. What would be the power dissipated in the circuit for an applied alternating voltage of frequency 1000 Hz and peak value 10 V?

Dear student assuming 40Ω resistance Z=XL2+R2=ω2L2+R2=4π2f2L2+R2=6280.12Ωcosθ=RZ=406280.12=0.0063I0=VZ=106280.12=0.0015P=V0I0cosθ2=10×0.0015×0.00632=0.0000047 WRegards