An LCR circuit contains resistance of $110 \Omega$
and a supply of $220 V$ at $300 rad / s$ angular
frequency. If only capacitance is removed from
the circuit, current lags behind the voltage by
$45^{\circ} .$ If on the other hand, only inductor is
removed the current leads by $45^{\circ}$ with the
applied voltage. The $rms$ current flowing in the circuit will be :
- $1 A$
- $2.5 A$
- $1.5 A$
- $2 A$
The Correct Option is D
Solution and Explanation
To determine the \(rms\) current flowing in the given LCR circuit, we need to analyze the conditions specified by the question step-by-step.
The circuit has a resistance of \(R = 110 \Omega\) and a supply voltage \(V = 220 \, \text{V}\) with an angular frequency of \(\omega = 300 \, \text{rad/s}\).
The problem states that the current lags by \(45^\circ\) when the capacitance is removed, which implies that the circuit contains only resistance and inductance. The phase angle \(\phi\) in an \(RL\) circuit is determined by:
\(\tan \phi = \frac{X_L}{R}\),
where \(X_L\) is the inductive reactance. Since \(\phi = 45^\circ\) and \(\tan 45^\circ = 1\), we have:
\(\frac{X_L}{R} = 1 \Rightarrow X_L = R = 110 \, \Omega\).
Substituting the inductive reactance formula \(X_L = \omega L\), we get:
\(110 = 300 \times L \Rightarrow L = \frac{110}{300} = \frac{11}{30} \, \text{H}\).
Similarly, when the inductor is removed, the current leads by \(45^\circ\), indicating a resistor-capacitor \((RC)\) circuit. Here, the phase angle is represented by:
\(\tan \phi = \frac{-X_C}{R}\),
where \(X_C\) is the capacitive reactance. Using \(\phi = 45^\circ\) and \(\tan 45^\circ = 1\), we have:
\(\frac{-X_C}{R} = 1 \Rightarrow X_C = -R = -110 \, \Omega\).
Therefore, \(X_C = \frac{1}{\omega C} = 110 \Rightarrow C = \frac{1}{300 \times 110} = \frac{1}{33000} \, \text{F}\).
The circuit behaves as an \({\text{RLC}}\) circuit, fully balanced because the inductive and capacitive reactances cancel each other out. In such a condition, the impedance \(Z\) of the circuit is purely resistive:
\(Z = R = 110 \, \Omega\).
Finally, we calculate the root mean square ( \(rms\)) current using the formula:
\(I_{\text{rms}} = \frac{V}{Z} = \frac{220}{110} = 2 \, \text{A}\).
Thus, the \(rms\) current flowing in the circuit is \(2 \, \text{A}\).
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main LCR Circuit Questions
- In series LCR circuit, the capacitance is changed from C to 4C. To keep the resonance frequency unchanged, the new inductance should be :
- Output of given circuit represents which logic gate :
- An AC voltage \( V = 20 \sin 200 \pi t \) is applied to a series LCR circuit which drives a current \( I = 10 \sin \left( 200 \pi t + \frac{\pi}{3} \right) \). The average power dissipated is:
- An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of 20 μF is .............. V.
- A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
Concepts Used:
LCR Circuit
An LCR circuit, also known as a resonant circuit, or an RLC circuit, is an electrical circuit consist of an inductor (L), capacitor (C) and resistor (R) connected in series or parallel.
Series LCR circuit
When a constant voltage source is connected across a resistor a current is induced in it. This current has a unique direction and flows from the negative to positive terminal. Magnitude of current remains constant.
Alternating current is the current if the direction of current through this resistor changes periodically. An AC generator or AC dynamo can be used as AC voltage source.
