Explanation
The resulting velocity in an LCR circuit is the sum of the velocities in inductance and capacitance then adding the remaining voltage with that of the resistance voltage.
The angle between the voltage source and current$$i$$gives the circuit phase angle.
The voltage triangle for the LCR circuit is given as
$${V_s} = \sqrt {V_R^2 + {{\left( {{V_L} - {V_c}} \right)}^2}} $$
The impedance of LCR circuit is given as,
$$Z = \sqrt {\left[ {{R^2} + {{\left( {\omega L - \frac{1}{{\omega C}}} \right)}^2}} \right]} $$
$$ = \sqrt {{{\left( {10} \right)}^2} + \left( {\left( {2 \times \pi \times 500 \times 8.1 \times {{10}^{ - 3}}} \right) - \dfrac{1}{{2 \times \pi \times 500 \times 12.5 \times {{10}^{ - 6}}}}} \right)} $$
$$ = 10\;{\rm{ohm}}$$
The rms value of the current is given as,
$${I_{rms}} = \dfrac{{{V_{rms}}}}{Z}$$
$$ = \frac{{100}}{{10}}$$
$$ = 10\;{\rm{A}}$$
The potential difference across the given resistor is given as,
$${V_R} = I \times R$$
$$ = 10 \times 10$$
$$ = 100\;{\rm{V}}$$
Thus, the potential difference across the resistor is $$100\;{\rm{V}}$$.
The impedance of the circuit is given as,
$$Z = \sqrt {{R^2} + {X_L}^2} $$
$$Z = \sqrt {{{\left( {88} \right)}^2} + {{\left( {2 \times \pi \times 50 \times 0.21} \right)}^2}} $$
$$Z = 110\;\Omega $$
The current in the circuit is given as,
$$I = \dfrac{V}{Z}$$
$$I = \dfrac{{220}}{{110}}$$
$$ = 2\;{\rm{A}}$$
The phase angle is given as,
$${\rm{\theta }} = {\tan ^{ - 1}}\left( {\dfrac{{{X_L}}}{R}} \right)$$
$${\rm{\theta }} = {\tan ^{ - 1}}\left( {\dfrac{{2 \times \pi \times 50 \times 0.21}}{{88}}} \right)$$
$$ = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$$
Thus, the current in the circuit is $$2\;{\rm{A}}$$ and the phase angle is $${\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$$.
In a series L-C-R circuit the voltage across the resistance , capacitance and inductance is 10 V each. If capacitance is short circuited, the voltage across the inductance will be:
The voltages across all the three components are equal, so the impedance will be the same.
$$R = {X_L}$$
The current is given as,
$$I = \dfrac{V}{{\sqrt {{R^2} + {{\left( {{X_L}} \right)}^2}} }}$$
$$I = \dfrac{{10}}{{R\sqrt 2 }}$$
The potential drop across inductor is given as,
$${V_L} = I{X_L}$$
$${V_L} = IR$$
$${V_L} = \dfrac{{10}}{{\sqrt 2 }}\;{\rm{V}}$$
The frequency of the resulting vibrations is given as,
$$f = \dfrac{1}{{2\pi \sqrt {LC} }}$$
$$f = \dfrac{1}{{2\pi \sqrt {10 \times {{10}^{ - 3}} \times 20 \times {{10}^{ - 6}}} }}$$
$$f = 356\;{\rm{cycle/s}}$$
Please disable the adBlock and continue. Thank you.