Question

A LCR series circuit with 100 ohm resistance is connected to an ac source of 200 V and angular frequency of 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 60 degrees. When only the inductor is removed, the current leads the voltage by 60 degrees. Calculate the current in the circuit.

Answer 1

Explanation:

HERE IS UR ANSWER.....

Answer 2

Answer:

The current in the circuit is equal to two amperes 2A.

Explanation:

When Capacitor is removed: Current lags behind the voltage by 60°

               $tan\phi =\frac{X_{L}}{R}$

Where $X_{L}$  is the reactance and R is resistance

              $tan60^{o} =\frac{X_{L}}{R}\\ {X_{L}=\sqrt{3} R$                               ..............(1)

When Inductor is removed: current leads the voltage by 60°

               $tan\phi =\frac{X_{C}}{R}$

where $X_{C}$ is reactance

              $tan60^{o}=\frac{X_{C}}{R}\\ X_{C}=\sqrt{3}R$                            .................(2)

From eq.(1) and (2)

             $X_{L}=X_{C}$                                  .................(3)

The above equation is the condition of resonance which means Z=R where Z is impedence.

We have given Z=R =100Ω and  E=200Volt

                     $I=\frac{E}{Z}\\ \\ I= \frac{200}{100} \\ I=2A$

Therefore the current in the given circuit is 2A.