A bulb and a capacitor are connected in series to a source of alternating current. If its frequency is increased, while keeping the voltage of the source constant, then:
As frequency increases, capacitive reactance \(X_C = \frac{1}{2\pi f C}\) decreases. This decreases the total impedance \(Z = \sqrt{R^2 + X_C^2}\), thereby increasing the current, causing the bulb to glow more brightly.
The phase difference between voltage and current in series LCR circuit at half power frequency is:
At half-power frequencies, the power factor is \(\cos\phi = \frac{1}{\sqrt{2}}\), which gives \(\phi = \pi/4\).
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.
A series LCR circuit containing \(5.0\text{ H}\) inductor, \(80 \mu\text{F}\) capacitor and \(40 \Omega\) resistor is connected to \(230\text{ V}\) variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be
The resonant angular frequency is \(\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5 \times 80 \times 10^{-6}}} = 50\text{ rad/s}\) and the bandwidth is \(Delta \omega = \frac{R}{L} = \frac{40}{5} = 8\text{ rad/s}\). The half-power frequencies are \(omega_0 \pm \frac{\Delta \omega}{2} = 50 \pm 4\text{ rad/s}\), which gives \(46\text{ rad/s}\) and \(54\text{ rad/s}\).
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 \[f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10^{-4} \times 10^{-6}}} = \frac{10^5}{2\pi}\text{ Hz}\].