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Series LR Circuit - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Series LR circuit is considered one the most difficult concept.

  • 46 Questions around this concept.

Solve by difficulty

 An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and then connected to the L and R as shown below :

If a student plots graphs of the square of maximum charge ( Q2Max ) on the capacitor with time(t) for two different values L1 and L2 (L 1>L2) of L then which of the following represents this graph correctly ? (plots are schematic and not drawn to scale)

 

The power factor of an  AC circuit having resistance (R)  and inductance (L)  connected in series and an angular velocity \omega is :

The phase difference between the alternating current and emf is \pi /2. Which of the following cannot be the constituent of the circuit?

 A series LR circuit is connected to a voltage source with  \mathrm{V}(\mathrm{t})=\mathrm{V}_0 \sin \omega \mathrm{t}.  After very large time, current  I(t)  behaves as

\left ( t_{0}\gg \frac{L}{R} \right ) :
 

Concepts Covered - 1

Series LR circuit

Series LR circuit-

                                               

The above figure shows that pure inductor of inductance L connected in series with a resistor of resistance R through sinusoidal voltage, which is given by -  \mathrm{V}=\mathrm{V}_{0} \sin (\omega \mathrm{t}+\varphi).

The alternating current I, which is flowing in the circuit gives rise to voltage drop VR across the resistor and voltage drop VL across the coil. As we have studied in previous concept that the voltage drop VR across R would be in phase with current but voltage drop across the inductor will lead the current by a phase factor π/2.

So, the voltage drop across R is - VR=IR

voltage drop across the inductor L is - \mathrm{V}_{\mathrm{L}}=\mathrm{I}(\omega \mathrm{L})

Where, I is the value of current in the circuit at a given instant of time

So, the voltage phasor diagram is - 

                                                        

In the above figure, we have taken current as a reference quantity because same amount of current flows through both the components. Thus from phasor diagram - 

                                                              \begin{aligned} V &=\sqrt{V_{R}^{2}+V_{L}^{2}} \\ \\ &=I \sqrt{R^{2}+\omega^{2} L^{2}} \\ \\ &=I Z \\ \text { where, } \\ Z &=\left(R^{2}+\omega^{2} L^{2}\right)^{1 / 2} \end{aligned}

Here, Z is known as Impedence of the circuit.

 

By using the formula of impedence we can write that - 

                                                                  I=\frac{V_{0} \sin (\omega t-\phi)}{Z}

This is current in steady state which lags behind applied voltage by an angle φ.

From here and the above figure, we can write that - 

                                                                              tan \varphi = \frac{\omega L}{R} = \frac{X_L}{R}

 

Important term - 

1. Power factor - 

                 \cos \phi = \frac{R}{Z}

                 R\rightarrow resistance

                 Z\rightarrow impedence

2. Inductive susceptance (S_{L}) - 

It is the reciprocal of reactance.

 

             S_{L}= \frac{1}{X_{L}}= \frac{1}{2\pi \nu L}

 

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Series LR circuit

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