A coil of resistance 300 - and inductance 1-0 H is connected across an alternating voltage of frequency 3002 - Hz- Calculate the phase difference between the voltage and current in the circuit-

Given: f = 3002 π Hz ⇒ω = 2π f = 300 rad/s Using phasor analysis, X_L= ω L = 300× 1=300 Ω So, tanθ = X_LR= 300 300 = 1 ⇒ nbsp; θ = 45^∘ So, phase di ...

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Byju's Answer

Standard XII

Physics

AC Applied to a Resistor

A coil of res...

Question

A coil of resistance $300 Ω$ and inductance $1.0 H$ is connected across an alternating voltage of frequency $\frac{300}{2 π} Hz$ . Calculate the phase difference between the voltage and current in the circuit.

\frac{π}{4}

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\frac{π}{3}

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π

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\frac{π}{6}

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Solution

The correct option is A $\frac{π}{4}$
Given: $f = \frac{300}{2 π} Hz$

$\Rightarrow ω = 2 π f = 300 rad/s$

Using phasor analysis,

$X_{L} = ω L = 300 \times 1 = 300 Ω$

So, $tan θ = \frac{X_{L}}{R} = \frac{300}{300} = 1$

$\Rightarrow θ = 45^{\circ}$

So, phase difference between current and voltage in circuit $= θ = 45^{\circ} = \frac{π}{4}$

Hence, option $(C)$ is correct.

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A coil of resistance 300 Ω and inductance 1.0 H is connected across an alternating voltage of frequency 3002π Hz. Calculate the phase difference between the voltage and current in the circuit.

The correct option is A π4 Given: f=3002π Hz ⇒ω=2πf=300 rad/s Using phasor analysis, XL=ωL=300×1=300 Ω So, tanθ=XLR=300300=1 ⇒θ=45∘ So, phase difference between current and voltage in circuit =θ=45∘=π4 Hence, option (C) is correct.

A coil of resistance $300 Ω$ and inductance $1.0 H$ is connected across an alternating voltage of frequency $\frac{300}{2 π} Hz$ . Calculate the phase difference between the voltage and current in the circuit.