A series L-C-R circuit is connected across an AC source E = `10sin[100pit - pi/6]`. Current from the supply is I = `2sin[100pit + pi/12]`, What is the average power dissipated?

A series L-C-R circuit is connected across an AC source V = 10sin[100pit - pi/6] . Current from the supply is I = 2sin[100pit + pi/12] , What is the average power dissipated?

A series LCR circuit is connected across a source of emf E = 20 sin (100pit - pi/6) . The current from the supply is I = 4sin (100pit + pi/12) . Draw the impedance triangle for the circuit.

A series LCR circuit is connected across a source of emf E = 20 sin (100pit - pi/6) . The current from the supply is I = 4sin (100pit + pi/12) . Draw the impedance triangle for the circuit.

A series LCR circuit is connected across a source of emf E =10 sin (100 pi l - pi /6) . The current from the supply si l = 2 sin (100 pi l + pi/12) . Draw the impedance triangle for the circuit.

A resistor and an inductor are connected in series to an AC source of voltage 150 sin (100 pit +pi) volt. If the current in the circuit is 5 sin (100 pit + (2pi)/(3)) ampere then the average power dissipated and the resistance of the resistor are respectively

In LCR series circuit , an alternating emf e and current i are given by the equations e = sin (100 t) volt . i=100 sin (100 t + (pi)/(3)) mA The average power dissipated in the circuit will be

In an a.c. circuit, voltage and current are given by : V = 100 sin (100 t) V and I = 100 sin (100 t (pi)/3) mA respectively. The average power dissipated in one cycle is :

An inductor 2//pi Henry, a capacitor 100//T muF and a resistor 75Omega are connected in series across a source of emf V=10sin 100pit . Here t is in second. (a) find the impedance of the circuit. (b) find the energy dissipated in the circuit in 20 minutes.