In an L-C-R circuit with all components connected in series,the e.m.f. and the current flowing in the circuit are given by the following equation : `e_d = 200 sin (314 .t + pi//6)` Volt,I = 5 sin 314 t A. obtain:The peak values of current and e.m.f.

In an L-C-R circuit with all components connected in series,the e.m.f. and the current flowing in the circuit are given by the following equation : e_d = 200 sin (314 .t + pi//6) Volt,I = 5 sin 314 t A. obtain: The phase difference between current and e.m.f.

In an L-C-R circuit with all components connected in series,the e.m.f. and the current flowing in the circuit are given by the following equation : e_d = 200 sin (314 .t + pi//6) Volt,I = 5 sin 314 t A. obtain: The frequency of the a.c. source.

The equation of an a.c. current I = 50 sin 100 pi t .Find effective value of current.

The e.m.f of and a.c. current E=158sin 200pit V . What is the voltage at time t=1/400 sec?

The instantaneous current in an ac circuit is I=6sin314tA . What is the rms value of current ?

In an ac circuit alternating voltage E= 200sqrt2 sin 100t volt is connected to a capacitor of capacity 1muF . The rms value of the current in the circuit is

In an LCR series combination, R=400 Omega L=100mH and C=1muF . This combination is connected to a 25sin2000t volt source. Find peak value of current,

An electric bulb is connected to battery of e.m.f. 10V and 0.01 A current is flowing through the bulb. But when it is connected to 220V mains then the cureent is 0.05 A. Explain this apparent discrepancy with ohm's law.

A resistor, R=300Omega and a capacitor , C=25muF are connected in series with an ac source. The peak value of voltage (V_0) and the frequency (f) of the source are 150 V and and 50/pi Hz respectively. Find the peak value of current and the power dissipated in the circuit.

A series combination of an inductor of self-inductance L, capacitor of capacitance C and resistor R is connected to an alternating voltage source of V=V_0sinomegat . The current through the circuit is I=I_0sin(omegat-theta) ,where I_0=V_0/(sqrt(R^2+(omega-1/(omegaC))^2)) and theta=tan^-1 ""1/R(omegaL-1/(omegaC)) . Now that, the frequency of both voltage and current is f=omega/(2pi) . The rms value of these parameters during one complete cycles are V_(rms)=V_0/sqrt2 and I_(rms)=I_0/sqrt2 respectively. These values are shown in alternating voltmeter and ammeter. The power consumed by the circuit P=VI. The mean value i.e., the effective power of the circuit in a complete cycle is overlineP=V_(rms)I_(rms)costheta . This costheta is termed the power factor. In the circuit in question (i) the inductor is replaced by a pure capacitor, the phase difference between the current and terminal voltage of the capacitor is