Exams › NEET › Physics

A resistance of 300 \(\Omega\) and an inductance of \(\frac{1}{\pi}\) henry are connected in series to an AC voltage of 20 V and 200 Hz frequency. The phase angle between the voltage and current is

1. \(\tan^{-1}\!\left(\frac{3}{4}\right)\)
2. \(\tan^{-1}\!\left(\frac{4}{3}\right)\)
3. \(\tan^{-1}\!\left(\frac{2}{3}\right)\)
4. \(\tan^{-1}\!\left(\frac{5}{2}\right)\)

Correct answer: \(\tan^{-1}\!\left(\frac{3}{4}\right)\)

Solution

In a series RL circuit, the phase angle satisfies \(\tan\phi = X_L/R\). Here \(X_L = 2\pi fL = 2\pi \times 200 \times \frac{1}{\pi} = 400\,\Omega\), so \(\tan\phi = 400/300 = 4/3\). The angle between voltage and current is therefore \(\tan^{-1}(4/3)\); since the current lags, the phase angle is often written with a negative sign depending on convention.

Related NEET Physics questions

• The current flowing through the resistor in a series LCR a.c. circuit, is I=ε / R Now the inductor and capacitor are connected in parallel and joined in series with the resistor as shown in figure. The current in the circuit is now. (Symbols have their usual meaning)
• Which of the following will not increase voltage in a simple generator?
• With increase in frequency of an A.C. supply, the inductive reactance
• ln an A C circuit, the instantanceous emf and current are given by e=100 sin 30 t, i=sin (30 t (π)/(4)) In one cycle of A C, the average power consumed by the circuit and the wattles current are, respectively
• The Joules heating effect for an electric heater in A. C and in D.C. circuit is
• A resistance (R)=12 Ω; inductance (L)=2 henry and capacitive reactance C=5 m F are connected in series to an ac generator, then:

⚔️ Practice NEET Physics free + battle 1v1 →