A coil of resistance \(20\Omega\) and self-inductance 10 mH is connected to an ac source of frequency \(1000/\pi \mathrm…
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A coil of resistance \(20\Omega\) and self-inductance 10 mH is connected to an ac source of frequency \(1000/\pi \mathrm{Hz}\). The phase difference between current in the circuit and the source voltage is :
✓ Correct answer: d)
\(45^\circ\)
ExplanationPlugging in the values for \(f\) and \(L\):
\({X}_{L}=2\pi (\frac{1000}{\pi })(10\times {10}^{-3})\)3 Sv6Kpe[] \({X}_{L}=20\ \Omega\) Calculate the phase difference (\(ϕ\)) The phase difference (\(ϕ\)) between the current and voltage in an RL circuit is given by the formula \(\tan (ϕ)=\frac{{X}_{L}}{R}\) \(\tan (ϕ)=\frac{20}{20}\) \(\tan (ϕ)=1\)1 Therefore, the phase difference is: \(ϕ=\mathrm{arctan}(1)={45}^{∘}\)or\(\frac{\pi }{4}\)radians.
\({X}_{L}=2\pi (\frac{1000}{\pi })(10\times {10}^{-3})\)3 Sv6Kpe[] \({X}_{L}=20\ \Omega\) Calculate the phase difference (\(ϕ\)) The phase difference (\(ϕ\)) between the current and voltage in an RL circuit is given by the formula \(\tan (ϕ)=\frac{{X}_{L}}{R}\) \(\tan (ϕ)=\frac{20}{20}\) \(\tan (ϕ)=1\)1 Therefore, the phase difference is: \(ϕ=\mathrm{arctan}(1)={45}^{∘}\)or\(\frac{\pi }{4}\)radians.
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