In an LRC series circuit at resonance,the cur | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Initially at resonance: $X_L = X_C \Rightarrow Z = R$.Therefore,the peak current is $I_0 = \frac{V_0}{R} = 10\sqrt{2} \ A$.When the current lags b

Vedclass · Accepted Answer

Frequency must be increased and current after the change is $10 \ A$.

Vedclass · Answer

(A) Initially at resonance: $X_L = X_C \Rightarrow Z = R$.Therefore,the peak current is $I_0 = \frac{V_0}{R} = 10\sqrt{2} \ A$.When the current lags by $45^o$,the circuit must be inductive,meaning $X_L > X_C$. This implies the frequency $\omega$ must be greater than the resonant frequency $\omega_0 = \frac{1}{\sqrt{LC}}$. Thus,the frequency must be increased.For a phase angle $\phi = 45^o$,we have $	an(45^o) = \frac{X_L - X_C}{R} = 1$,which means $X_L - X_C = R$.The new impedance is $Z' = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + R^2} = R\sqrt{2}$.The new current is $I' = \frac{V_0}{Z'} = \frac{V_0}{R\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{2}} = 10 \ A$.

In an $LRC$ series circuit at resonance,the current in the circuit is $10\sqrt{2} \ A$. If the frequency of the source is changed such that the current now lags by $45^o$ behind the applied voltage,which of the following is correct?

Similar Questions

Which graph shows the correct variation of r.m.s. current '$i$' with frequency '$f$' of a.c. in case of a series resonant circuit?

Power delivered by the $ac$ source of the circuit becomes maximum when

$A$ resistor of $2 \ \Omega$,an inductor of $100 \ \mu H$,and a capacitor of $400 \ pF$ are connected in series across an $A$.$C$. source of $e_{rms} = 0.1 \ V$. At resonance,the voltage drop across the inductor is: (in $V$)

An $LCR$ series circuit is connected to an external $emf$,$e = 200 \sin(100 \pi t) \ V$. The values of capacitance and resistance in the circuit are $1 \ \mu F$ and $100 \ \Omega$ respectively. The amplitude of current in the circuit is maximum when the inductance is (in henry):

Vedclass Test Series

Exam Paper Generator

Online Exam Module