Alternating Current MCQ Questions & Answers in Electrostatics and Magnetism | Physics

Impedance at resonant frequency is minimum in series $$LCR$$  circuit.
So, $$Z = \sqrt {{R^2} + {{\left( {2\pi fL - \frac{1}{{2\pi fC}}} \right)}^2}} $$
When frequency is increased or decreased, $$Z$$ increases.

According to given problem,
$$\eqalign{ & I = \frac{V}{Z} = \frac{V}{{{{\left[ {{R^2} + {{\left( {\frac{1}{{C\omega }}} \right)}^2}} \right]}^{\frac{1}{2}}}}}\,......\left( {\text{i}} \right) \cr & {\text{and}}\,\,\frac{I}{2} = \frac{V}{{{{\left[ {{R^2} + {{\left( {\frac{3}{{C\omega }}} \right)}^2}} \right]}^{\frac{1}{2}}}}}\,......\left( {{\text{ii}}} \right) \cr} $$
Substituting the value of $$I$$ from eq. (i) in (ii),
$$\eqalign{ & 4\left( {{R^2} + \frac{1}{{{C^2}{\omega ^2}}}} \right) = {R^2} + \frac{9}{{{C^2}{\omega ^2}}} \cr & {\text{i}}{\text{.e}}{\text{.,}}\,\frac{1}{{{C^2}{\omega ^2}}} = \frac{3}{5}{R^2} \cr & {\text{So}}\,{\text{that}}\,\,\frac{X}{R} \cr & = \frac{{\left( {\frac{1}{{C\omega }}} \right)}}{R} = \frac{{{{\left( {\frac{3}{5}{R^2}} \right)}^{\frac{1}{2}}}}}{R} = \sqrt {\frac{3}{5}} \cr} $$

$$\eqalign{ & {\text{Here,}}\,{X_L} = {\omega _L} = 2\pi fL = 2\pi \times 50 \times 1 = 100\,\pi \Omega \cr & {X_c} = \frac{1}{{\omega C}} = \frac{1}{{2\pi fC}} = \frac{1}{{2\pi \times 50 \times 10 \times {{10}^{ - 6}}}} = \frac{{{{10}^3}}}{\pi }\Omega \cr & {\text{So,}}\,X = \left| {{X_L} - {X_C}} \right| = \left| {100\pi - \frac{{{{10}^3}}}{\pi }} \right| = \left| {{{10}^2}\left[ {\frac{{{\pi ^2} - 10}}{\pi }} \right]} \right|\Omega \cr} $$

$$I = {I_C} + \left( { - {I_L}} \right)$$
Because $${I_C}$$ and $${I_L}$$ are $${180^ \circ }$$ out of phase with each other.

$$\cos \phi = \frac{R}{Z},$$   where $$Z$$ is the impedance & $$Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} ,$$      if there is only resistance then $$Z = R \Rightarrow \cos \phi = 1$$

$$\eqalign{ & I\left( 0 \right) = \frac{{15 \times 100}}{{0.15 \times {{10}^3}}} = 0.1\,A \cr & I\left( \infty \right) = 0 \cr & I\left( t \right) = \left[ {I\left( 0 \right) - I\left( \infty \right)} \right]{e^{\frac{{ - t}}{{\frac{L}{T}}}}} + i\left( \infty \right) \cr & I\left( t \right) = 0.1\,{e^{\frac{{ - t}}{{\frac{L}{R}}}}} = 0.1\,{e^{\frac{R}{L}}} \cr & I\left( t \right) = 0.1\,e\frac{{0.15 \times 1000}}{{0.03}} = 0.67\,mA \cr} $$

In series $$LCR$$  circuit at resonance,
$${\text{capacitive reactance}}\left( {{X_C}} \right) = {\text{inductive reactance}}\left( {{X_L}} \right)$$
$${\text{i}}{\text{.e}}{\text{.}}\,\,\frac{1}{{\omega C}} = \omega L$$
Total impedance of the circuit
$$\eqalign{ & Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \cr & = \sqrt {{R^2} + {{\left( {\omega L - \frac{1}{{\omega C}}} \right)}^2}} \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,Z = R \cr} $$
So, power factor, $$\cos \phi = \frac{R}{Z} = \frac{R}{R} = 1$$
Thus, power loss at resonance is given by
$$\eqalign{ & P = {V_{rms}}\,{i_{rms}}\cos \phi \,\,\left[ {\cos \phi = 1} \right] \cr & = {V_{rms}}\,{i_{rms}} \times 1 \cr & = \left( {{i_{rms}}R} \right){i_{rms}} \cr & = {\left( {{i_{rms}}} \right)^2}R = {i^2}R \cr} $$

Phase difference for $$R - L$$  circuit lies between $$\left( {0,\frac{\pi }{2}} \right)$$

$$\eqalign{ & {U_e} + {U_m} = 160 \cr & \therefore {U_m} = 160 - {U_e} = 160 - 10 \cr & = 150\,\mu J. \cr} $$

Time constant is $$\frac{L}{R}$$
Given, $$L=40\,H$$
$$\eqalign{ & \& R = 8\Omega \cr & \therefore \tau = \frac{{40}}{8} = 5\,\sec . \cr} $$