Question

In LCR circuit at resonance, current in the circuit is $10\sqrt 2 A$. If now the frequency of the source is changed such that current lags by${45^0}$ than the applied voltage in the circuit then, which of the following is correct?
$\left( {\text{A}} \right)$ Frequency must be increased and current after the change is \[{\text{10A}}\].
$\left( {\text{B}} \right)$ Frequency must be decreased and current after the change is \[{\text{10A}}\].
$\left( {\text{C}} \right)$ Frequency must be decreased and current is the same as that of initial value.
$\left( {\text{D}} \right)$ The given information is insufficient to conclude anything.

Answer 1

Hint:The resonance phenomenon is exhibited by a circuit only if both \[{\text{L}}\] and \[{\text{C}}\] are present in the circuit. Only then, the voltage across \[{\text{L}}\] and \[{\text{C}}\] cancel each other so that the voltage across \[{\text{R}}\] is equal to the source voltage.

Complete step by step answer:
Consider a circuit containing an inductor, capacitor and resistor are connected in series across an alternating source of voltage \[{\text{V}}\] or emf $\varepsilon $ .

Let the source supplies a sinusoidal voltage which is given by,
$V = {V_0}\sin \omega t$
Where, ${V_0}$ is the peak value of voltage $\omega $ is the angular frequency and \[t\] is the time period.
Let \[q\] be the charge on the capacitor and \[i\] be the current in the circuit at any instant of time \[t\] .
Let ${V_R},{V_L},{V_C}$ represent the voltage across the resistor, inductor, and capacitor respectively.
Then, the voltage across the resistor, ${V_R} = {i_0}R$
The voltage across an inductor, ${V_L} = {i_0}{X_L}$
The voltage across a capacitor, ${V_C} = {i_0}{X_C}$
Where, ${i_0}$ is the peak value of current, ${X_C}$ is capacitive reactance, ${X_L}$ is the inductive reactance, and \[{\text{R}}\] is the resistance of the resistor.
It is stated in the questions that are the current lags by${45^0}$ that is,$\phi = {45^0}$
Then, $\cos \phi = \dfrac{R}{Z}$
$\cos {45^0} = \dfrac{R}{Z}$
Here $\cos {45^0} = \dfrac{1}{{\sqrt 2 }}$
So we can write it as,
$\dfrac{1}{{\sqrt 2 }} = \dfrac{R}{Z}$
Taking cross multiplication we get,
$ \Rightarrow Z = \sqrt 2 R$
At resonance condition,${X_L} = {X_C}$
Then impedance $Z = \sqrt {\left( {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \right)} $ becomes, $Z = R$
There is a maximum value of current is flowing in the circuit ${I_{rms}} = 10\sqrt 2 A$
That is, ${I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$
$10\sqrt 2 = \dfrac{{{I_0}}}{{\sqrt 2 }}$
After simplification we get,
${I_0} = 10A$
Now we already have an given that, current lags behind the applied voltage by \[\;45\] degree, then ${X_L} > {X_C}$ for this condition angular frequency $\omega $ must be increases, so that ${X_L}$ will increase and ${X_C}$ will decrease.
Therefore, the frequency must be increased and current after the change is \[{\text{10A}}\].

The correct option is A

Note:
Inductive reactance is the opposition offered by an inductance to the ac through it.
Capacitive reactance is the effective opposition offered by the capacitance to the ac through it.
Impedance of an ace circuit is the effective opposition offered by the circuit to the alternating current through it.