An inductor of inductance Land resistor of resistance Rare connected in series to an AC source of voltage V = V0 sin(omega t). The current in the circuit is I = I0 sin(omega t - phi). The value of tan(phi) is:

A. L/R — In an LR series circuit, the voltage leads the current by a phase angle phi where tan(phi) is the ratio of inductive reactance to resistance. Thus, tan(phi) = omega L / R.

JEE Main · PhysicsHard

$An inductor of inductance L and resistor of resistance R are connected in series to an AC source of voltage V = V_{0} sin (ω t) . The current in the circuit is I = I_{0} sin (ω t - ϕ) . The value of tan (ϕ) is:$

A. $ω L / R$
B. $R / (ω L)$
C. $ω R / L$
D. $L / (ω R)$

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Correct answer: A — $ω L / R$

Solution

In an LR series circuit, the voltage leads the current by a phase angle ϕ where tan (ϕ) = X_{L} / R . Substituting X_{L} = ω L, we get tan (ϕ) = ω L / R .

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An inductor of inductance L and resistor of resistance R are connected in series to an AC source of voltage V=V0​sin(ωt). The current in the circuit is I=I0​sin(ωt−ϕ). The value of tan(ϕ) is:

Solution

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More AC Circuits practice questions

$An inductor of inductance L and resistor of resistance R are connected in series to an AC source of voltage V = V_{0} sin (ω t) . The current in the circuit is I = I_{0} sin (ω t - ϕ) . The value of tan (ϕ) is:$