Question

A series LCR circuit containing a resistance of 120Ω has angular frequency $4 \times {10^5}$rad / s . At resonance the voltages across resistance and inductance are 60 V and 40 V respectively. Find inductance and the frequency at which the current in the circuit lags the voltage by ${45^0}$ is:
A. $2 \times {10^{ - 4}}H;8 \times {10^5}Hz$
B. $2 \times {10^{ - 4}}H;16 \times {10^5}Hz$
C. $2 \times {10^{ - 4}}H;24 \times {10^5}Hz$
D. None of these.

Answer 1

In case of DC circuit if there is resistor we can measure using an ammeter but if there are inductors and capacitors in the AC circuit we can’t measure the obstruction for the flow of current hence there comes a term inductive reactance and capacitive reactance and they depend on source angular frequency too.

Formula used:
\eqalign{ & {X_C} = \dfrac{1}{{\omega C}} \cr & {X_L} = \omega L \cr & \cos \phi = \dfrac{R}{Z} \cr & Z = \sqrt {{{({X_L} - {X_C})}^2} + {R^2}} \cr}

Complete step by step answer:
In case alternating currents there will be angular velocity of a wave and we can find frequency of an alternating wave from that. That frequency determines the capacitive reactance and inductive reactance and total impedance of the AC circuit. If a capacitor is present in the AC circuit then we include the capacitive reactance and if not then only resistance and inductor reactance determines the impedance.
Normally AC voltages will be in the form $V = {V_0}\sin (\omega t)$ where ${V_0}$ is the peak voltage while $\omega$ is the angular frequency.
In AC circuits there will be some phase difference between the current and the potential in the circuit. That phase difference is required to be ${45^0}$.
Let current passing be ‘i’
From ohm’s law
\eqalign{ & {V_R} = iR \cr & \Rightarrow 60 = i\left( {120} \right) \cr & \therefore i = 0.5A \cr}
Inductor reactance will be given as ${X_L} = \omega L$
\eqalign{ & {V_L} = i{X_L} \cr & \Rightarrow i = 0.5A \cr & \Rightarrow 40 = 0.5\left( {\omega L} \right) \cr & \Rightarrow 40 = 0.5\left( {4 \times {{10}^5} \times L} \right) \cr & \therefore L = 2 \times {10^{ - 4}}H \cr}
At resonance condition voltage across the capacitor and inductor will be the same.
\eqalign{ & {V_L} = {V_C} \cr & \Rightarrow i{X_L} = i{X_C} \cr & \Rightarrow {X_L} = {X_C} \cr & \Rightarrow \omega L = \dfrac{1}{{\omega C}} \cr & \Rightarrow C = \dfrac{1}{{{\omega ^2}L}} \cr & \Rightarrow C = \dfrac{1}{{{{(4 \times {{10}^5})}^2}(2 \times {{10}^{ - 4}}H)}} \cr & \therefore C = 3.125 \times {10^{ - 8}}F \cr}
The phase difference($\phi$) is required to be 45 degrees. We have
$\cos \phi = \dfrac{R}{Z}$
Where Z is an impedance and given by
$Z = \sqrt {{{({X_L} - {X_C})}^2} + {R^2}}$
\eqalign{ & \Rightarrow \cos \phi = \dfrac{R}{Z} \cr & \Rightarrow \cos 45 = \dfrac{R}{Z} \cr & \Rightarrow \dfrac{1}{{\sqrt 2 }} = \dfrac{R}{Z} \cr & \Rightarrow Z = \sqrt 2 R \cr & \Rightarrow Z = \sqrt {{{({X_L} - {X_C})}^2} + {R^2}} \cr & \Rightarrow \sqrt {{{({X_L} - {X_C})}^2} + {R^2}} = \sqrt 2 R \cr & \Rightarrow {({X_L} - {X_C})^2} = {R^2} \cr & \Rightarrow {X_L} - {X_C} = R \cr & \Rightarrow \omega (2 \times {10^{ - 4}}H) - \dfrac{1}{{\omega (3.125 \times {{10}^{ - 8}}F)}} = 120 \cr & \Rightarrow \omega = 8 \times {10^5}rad/\sec \cr & \Rightarrow 2\pi f = 8 \times {10^5}rad/\sec \cr & \therefore f = 1.273 \times {10^5}Hz \cr}
None of the options has the given frequency value.
So option D is correct.

Note:
Here the magnitude of angular frequency we got is there in the one of the options. But we want the frequency, not angular frequency. Units of frequency are cycles per second or hertz while the units of angular frequency are radians per second. Both are different. Hence none is the answer.