The peak voltage of the AC source is equal to
For a sinusoidal AC voltage source, the relation between peak voltage \(V_0\) and root-mean-square voltage \(V_{\text{rms}}\) is \(V_0 = \sqrt{2}V_{\text{rms}}\).
The peak voltage of the AC source is equal to
For a sinusoidal AC voltage source, the relation between peak voltage \(V_0\) and root-mean-square voltage \(V_{\text{rms}}\) is \(V_0 = \sqrt{2}V_{\text{rms}}\).
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 \[f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10^{-4} \times 10^{-6}}} = \frac{10^5}{2\pi}\text{ Hz}\].
Power consumed in A.C circuit is zero then the ac source could be connected to
The average power consumed in an AC circuit is given by \( P_{avg} = V_{rms} I_{rms} \cos \phi \). For a purely inductive or purely capacitive circuit, the phase difference \( \phi = 90^\circ \), which makes the power factor \( \cos \phi = 0 \), resulting in zero power consumption.
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.
Assertion (A): ac current flows through a bulb and a solenoid connected in series. If a soft iron core is inserted in the solenoid, the bulb glows much brighter.
Reason (R): The inductance of solenoid decreases on inserting soft iron core in it.
Inserting a soft iron core increases solenoid inductance \(L\), thus increasing inductive reactance \(X_L = omega L\). This increases circuit impedance \(Z\), reducing current \(I = V/Z\) and making the bulb dimmer. Both Assertion (A) and Reason (R) are false.
Assertion (A): A choke coil has the characteristic of high inductance and low resistance.
Reason (R): More is the inductive property of the choke coil, Power factor of the circuit approaches maximum.
A choke coil has high inductance and low resistance (A is true). Power factor is \(cos\phi = R/Z = R/sqrt{R^2 + X_L^2}\). Higher inductive property (large \(X_L\)) makes \(cos\phi\) approach minimum (0), not maximum. So R is false.
Assertion (A): In a series \(LCR\) circuit at resonance, the voltage across the capacitor or inductor may be more than the applied voltage.
Reason (R): At resonance in a series \(LCR\) circuit, the voltages across inductor and capacitor are out of phase.
At resonance, \(V_L = V_C\) but they are \(180^\circ\) out of phase. The applied voltage is \(V = IR\). Due to voltage magnification (high \(Q\) factor), \(V_L\) or \(V_C\) can be much greater than \(V\). Reason is true, but it doesn't explain *why* they can be larger than applied voltage, it explains why they cancel out to make \(V=IR\).
Assertion (A): Average power consumed in an \(AC\) circuit is equal to average power consumed by resistors in the circuit.
Reason (R): Average power consumed by capacitor and inductor is zero.
Average power in \(AC\) is \(P_{avg} = V_{rms} I_{rms} \cos\phi\). For pure inductor or capacitor, \(\phi = \pm \pi/2\) so \(cos\phi = 0\). Only resistors dissipate average power, \(P_{avg} = I_{rms}^2 R\). Hence, R correctly explains A.