An inductor 20 mH, a capacitor 50 μF and a resistor 40 Ω are connected in series across a source of emf V = 10 sin340t. The power loss in the AC circuit is

A small signal voltage V(t) = V₀ sinωt is applied across an ideal capacitor C. What is the behavior of the current I(t)?

The magnetic potential energy stored in a certain inductor is $25$ mJ, when the current in the inductor is $60$ mA. This inductor is of inductance

An inductor $20$ mH, a capacitor $100 \mu F$ and a resistor $50 \Omega$ are connected in series across a source of emf, $V = 10 \sin 314 t.$ The power loss in the circuit is

A $40 \mu$F capacitor is connected to a $200$ V, $50$ Hz AC supply. The rms value of the current in the circuit is, nearly:

A series LCR circuit is connected to an AC voltage source. When L is removed from the circuit, the phase difference between current and voltage is $\frac{\pi}{3}.$ If instead C is removed from the circuit, the phase difference is again $\frac{\pi}{3}$ between current and voltage. The power factor of the circuit is:

A capacitor of capacitance $C$ is connected across an ac source of voltage $V$, given by$\newline V = V_0 \sin \omega t.$ The displacement current between the plates of the capacitor would then be given by:

An inductor of inductance $L,$ a capacitor of capacitance $C$ and a resistor of resistance $R$ are connected in series to an ac source of potential difference $V$ volts as shown in figure.$\newline$Potential difference across $L, C$ and $R$ is $40 V, 10 V$ and $40 V,$ respectively. The amplitude of current flowing through the LCR series circuit is $10 \sqrt{2} A.$ The impedance of the circuit is:

A series LCR circuit containing $5.0 H$ inductor, $80 \mu F$ capacitor, and $40 \Omega$ resistor is connected to a $230$ 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:

A step-down transformer connected to an AC mains supply of $220$ V is made to operate at $11$ V, $44$ W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?

The peak voltage of the AC source is equal to:

A series LCR circuit with inductance $10H,$ capacitance $10 \mu F,$ resistance $50 \Omega$ is connected to an AC source of voltage, $V = 200 \sin(100t)$ volt. If the resonant frequency of the LCR circuit is $v_0$ and the frequency of the AC source is $v,$ then:

A 12 V, 60 W lamp is connected to the secondary of a step-down transformer, whose primary is connected to ac mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?

An AC source is connected to a capacitor C. Due to decrease in its operating frequency:

In a series LCR circuit, the inductance L is 10 mH, capacitance C is $1\mu F$ and resistance R is $100 \Omega.$ The frequency at which resonance occurs is:

The net impedance of circuit (as shown in figure) will be:

In an ideal transformer, the turns ratio is $\frac{N_p}{N_s} = \frac{1}{2}.$ The ratio $\frac{V_s}{V_p}$ is equal to (symbols carry their usual meaning):

A $10 \mu F$ capacitor is connected to a $210 V, 50 Hz$ source as shown in figure. The peak current in the circuit is nearly ( $\pi = 3.14$):