For a series RLC circuit \(R = X_L = 2X_C\). The impedance of the circuit and phase difference between \(V\) and \(i\) will be:
Impedance \(Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (R - R/2)^2} = \frac{\sqrt{5}R}{2}\). Phase difference \(\tan\phi = \frac{X_L - X_C}{R} = 1/2 \Rightarrow \phi = \tan^{-1}(1/2)\).
An L-C-R series circuit with \(100\ \Omega\) resistance is connected to an ac source of \(100\text{ V}\) and angular frequency \(300\text{ rad/s}\). When the capacitance is removed, the current lags behind the voltage by \(45^\circ\). When the inductance is removed, the current leads the voltage by \(45^\circ\). The current flowing in the original circuit will be:
Removing C gives \(X_L = R\) and removing L gives \(X_C = R\). Hence, \(X_L = X_C\), indicating resonance. Impedance at resonance is \(Z = R = 100\ \Omega\). Current \(I = V/Z = 100/100 = 1\text{ A}\).
In a circuit \(L\), \(C\) and \(R\) are connected in series with an alternating voltage source of frequency \(f\). If the current lags the voltage by \(45^\circ\), then the value of \(L\) will be
Since current lags voltage, \(tan(45^\circ) = \frac{X_L - X_C}{R} = 1 ⇒ X_L - X_C = R\). Substituting \(X_L = 2\pi f L\) and \(X_C = \frac{1}{2\pi f C}\), we get \(2\pi f L = R + \frac{1}{2\pi f C} = \frac{2\pi f C R + 1}{2\pi f C}\). Thus, \(L = \frac{2\pi f C R + 1}{4\pi^2 f^2 C}\).
Statement A: In a pure capacitive ac circuit, the phase difference between current and voltage is \(\pi/2\).
Statement B: In a pure inductive ac circuit, the phase difference between current and voltage is \(\pi\).
In a purely capacitive circuit, current leads the voltage by \(\pi/2\). In a purely inductive circuit, current lags the voltage by \(\pi/2\). Hence, Statement A is correct while Statement B is incorrect.
The value of capacitance \(C\) in an LCR series circuit with inductance \(L = 50\) mH, connected with a voltage source of angular frequency \(100\) rad/s so that current in the circuit becomes maximum, is:
Current is maximum at resonance, where \(\omega = \frac{1}{\sqrt{LC}}\). Solving for \(C\): \(C = \frac{1}{\omega^2 L} = \frac{1}{(100)^2 \times 50 \times 10^{-3}} = 2 \times 10^{-3}\)⇒ C = 2 mF.
Consider the following statements:
A. Average power supplied to an inductor by an ac source over one complete cycle is zero.
B. Capacitive reactance is inversely proportional to both frequency of ac source and capacitance.
C. Power factor of AC series LCR circuit is equal to the product of resistance and impedance of the circuit.
The correct statement(s) is/are
Statement A is correct as average power of an inductor over a cycle is zero. Statement B is correct because \(X_C = \frac{1}{2pi f C}\). Statement C is incorrect because power factor is the ratio of resistance to impedance, \(cos \phi = R/Z\).
A series LCR circuit contains a capacitor of capacitance \(10^{-6}\text{ F}\) and an inductor of inductance \(10^{-4}\text{ H}\). The resonant frequency of the circuit will be
The resonant frequency is \[f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10^{-4} \times 10^{-6}}} = \frac{10^5}{2\pi}\text{ Hz}\].
Consider an a.c circuit having resistor of resistance \( R \) and an inductor having reactance \( X_L \). If voltage leads the current by an angle \( 60^\circ \), then
In an inductive-resistive (RL) series circuit, the phase angle \( \phi \) is given by \( \tan \phi = \frac{X_L}{R} \). Since \( \phi = 60^\circ \), \( \tan 60^\circ = \frac{X_L}{R} \implies X_L = \sqrt{3}R \).
A series LCR circuit contains a capacitor of capacitance \(10^{-6}\text{ F}\) and an inductor of inductance \(10^{-4}\text{ H}\). The resonant frequency of the circuit will be
The resonant frequency of a series LCR circuit is given by \(f_0 = \frac{1}{2\pi\sqrt{LC}}\). Substituting values: \(f_0 = \frac{1}{2\pi\sqrt{10^{-4} \times 10^{-6}}} = \frac{10^5}{2\pi}\text{ Hz}\).
Assertion (A): If an iron rod is inserted into a steady current carrying solenoid, the current in solenoid decreases.
Reason (R): Magnetic flux linked with solenoid increases.
Inserting an iron rod increases magnetic flux \( \Phi \). By Lenz's law, this induces an opposing \( \text{EMF} \), causing current \( \text{I} \) to decrease during insertion.