## LESSON 16. Sinusoidal Response of R-L -C Circuit

16.1. Sinusoidal Response of R-L-C Circuit

Consider a circuit consisting of resistance, inductance and capacitance in series as shown in Fig.16.1.  Switch S is closed at t = 0.  At t = 0, a sinusoidal voltage V cos (wt+q) is applied to the RLC series circuit, where V is the amplitude of the wave and q is the phase angle.  Application of Kirchhoff’s voltage law to the circuit results in the following differential equation.

$V\;\cos \,\left( {\omega t + \theta } \right)=Ri + L{{di} \over {dt}} + {1 \over C}\int {idt} ...................................................\left( {16.1} \right)$

Fig16.1

Differentiating the above equation, we get

$R{{di} \over {dt}} + L{{{d^2}i} \over {d{t^2}}} + i/C=-V\omega \,\sin \,\left( {\omega t + \theta } \right)$

$\left( {{D^2} + {R \over L}D + {1 \over {LC}}} \right)i=-{{V\omega } \over L}\,\sin \,\left( {\omega t + \theta } \right)...................................................\left( {16.2} \right)$

The particular solution can be obtained by using undermined coefficients.  By assuming

${i_p}=A\,\cos \,\left( {\omega t + \theta } \right) + B\,\sin \,\left( {\omega t + \theta } \right)...................................................\left( {16.3} \right)$

$i_p^'=A\omega \,\sin \,\left( {\omega t + \theta } \right) + B\omega \,\cos \,\left( {\omega t + \theta } \right)...................................................\left( {16.4} \right)$

$i_p^{''}=A{\omega ^2}\,\cos \,\left( {\omega t + \theta } \right) + B{\omega ^2}\,\sin \,\left( {\omega t + \theta } \right)...................................................\left( {16.5} \right)$

Substituting ip i'p and i’’p in Eq.16.2, we have

$\left\{ { - A\omega 2\,\cos \,\left( {\omega t + \theta } \right) + B\,\sin \,\left( {\omega t + \theta } \right)} \right\} + {R \over L}\left\{ { - A\omega \,\sin \,\left( {\omega t + \theta } \right) + B\omega \,\cos \,\left( {\omega t + \theta } \right)} \right\}$

$+ {1 \over {LC}}\left\{ {A\,\cos \,\left( {\omega t + \theta } \right) + B\,\sin \,\left( {\omega t + \theta } \right)} \right\}=-{{V\omega } \over L}\,\sin \left( {\omega t + \theta } \right)...................................................\left( {16.6} \right)$

Comparing both sides, we have

Sine coefficients

$-B{\omega ^2} - A{{\omega R} \over L} + {B \over {LC}}=-{{V\omega } \over L}$

$A\left( {{{\omega R} \over L}} \right) + B\left( {{\omega ^2} - {1 \over {LC}}} \right)={{V\omega } \over L}...................................................\left( {16.7} \right)$

Cosine coefficients

$- A{\omega ^2} + B{{\omega R} \over L} + {A \over {LC}}=0$

$A\left( {{\omega ^2} - {1 \over {LC}}} \right) - B\left( {{{\omega R} \over L}} \right)=0...................................................\left( {16.8} \right)$

Solving Eqs 16.7 and 16.8, we get

$A =\frac{{V\times\frac{{{\omega^2}R}}{{{L^2}}}}}{{\left[{{{\left({\frac{{\omega R}}{L}}\right)}^2}-{{\left({{\omega ^2}=\frac{1}{{LC}}}\right)}^2}}\right]}}$

$B=\frac{{\left({{\omega^2}-\frac{1}{{LC}}}\right)V\omega}}{{L\left[{{{\left({\frac{{\omega R}}{L}}\right)}^2}-{{\left({{\omega^2}=\frac{1}{{LC}}}\right)}^2}}\right]}}$

Substituting the values of A and B in Eq. 6.3, we get

${i_p}=\frac{{V\frac{{{\omega ^2}R}}{{{L^2}}}}}{{\left[{{{\left({\frac{{\omega R}}{L}}\right)}^2}-{{\left({{\omega ^2}-\frac{1}{{LC}}}\right)}^2}}\right]}}\cos\,\left({\omega t+\theta}\right)$

$+\frac{{\left({{\omega ^2}-\frac{1}{{LC}}}\right)V\omega}}{{L\left[{{{\left({\frac{{\omega R}}{L}}\right)}^2}-{{\left({{\omega ^2}-\frac{1}{{LC}}}\right)}^2}}\right]}}\sin\,\left({\omega t +\theta}\right)................................\left( {16.9} \right)$

Putting $M\,\cos\,\varphi=\frac{{V\frac{{{\omega^2}R}}{{{L^2}}}}}{{{{\left({\frac{{\omega R}}{L}} \right)}^2}-{{\left({{\omega ^2}-\frac{1}{{LC}}}\right)}^2}}}$

and $M\,\sin\,\varphi=\frac{{V\left({\omega 2-\frac{1}{{LC}}}\right)\omega}}{{L\left[{{{\left({\frac{{\omega R}}{L}}\right)}^2}-{{\left({{\omega ^2}-\frac{1}{{LC}}}\right)}^2}}\right]}}$

To find M and ø we divide one equation by the other

or ${{M\,\sin \,\phi} \over {M\,\cos \,\phi }}=\tan \,\phi={{\left( {\omega L - {1 \over {\omega C}}} \right)} \over R}$

$\phi=\tan-1\left[ {\left( {\omega L - {1 \over {\omega C}}} \right)/R} \right]$

Squaring both equations and adding, we get

${M^2}\,{\cos ^2}\,\varphi \,+\,{M^2}\,{\sin ^2}\,\varphi \,=\frac{{{V^2}}}{{{R^2}+{{\left({\frac{1}{{\omega C}}-\omega L}\right)}^2}}}$

$M=\frac{V}{{\sqrt{{R^2}+{{\left({\frac{1}{{\omega C}}-\omega L}\right)}^2}}}}$

The particular current becomes

${i_p}=\frac{V}{{\sqrt {{R^2}+{{\left({\frac{1}{{\omega C}}-\omega L} \right)}^2}}}}\cos\left[{\omega t+\theta+{{\tan}^{-1}}\frac{{\left({\frac{1}{{\omega C}}-\omega L}\right)}}{R}}\right].................................................\left( {16.10} \right)$

The complementary function is similar to that of DC series RLC circuit.  To find out the complementary function, we have the characteristic equation

$\left( {{D^2} + {R \over L}D + {1 \over {LC}}} \right)=0...................................................\left( {16.11} \right)$

The roots of Eq.16.11, are

${D_1},{D_2}={{ - R} \over {2L}} \pm \sqrt {{{\left( {{R \over {2L}}} \right)}^2} - {1 \over {LC}}}$

By assuming ${K_1}=-{R \over {2L}}and\,{K_2}=\sqrt {{{\left( {{R \over {2L}}} \right)}^2} - {1 \over {LC}}}$

${D_1}={K_1} + {K_2}\,\,and\,\,{D_2}={K_1} - {K_2}$

K2 becomes positive, when (R/2L)2>1/LC

The roots are real and unequal, which gives an overdamped response.  Then Eq.16.11 becomes

$\left[ {D - \left( {{K_1} + {K_2}} \right)} \right]\,\,\left[ {D - {K_1} - {K_2}} \right]i=0$

The complementary function for the above equation is

${i_c}={c_1}{e^{\left( {K1 + K2} \right)t}} + {c_2}{e^{\left( {K1 - K2} \right)t}}$

Therefore, the complete solution is

$i={i_c} + {i_p}$

${i_c}={c_1}{e^{\left({K1+K2}\right)t}}+{c_2}{e^{\left({K1-K2}\right)t}}+\frac{V}{{\sqrt{{R^2}+{{\left({\frac{1}{{\omega C}}-\omega L}\right)}^2}}}}\cos\,\left[{\omega t +\theta+{{\tan}^{-1}}\left({\frac{1}{{\omega CR}}-\frac{{\omega L}}{R}} \right)} \right]$

K2 becomes negative, when ${\left( {{R \over {2L}}} \right)^2} < {1 \over {LC}}$

Then the roots are complex conjugate, which gives an under damped response, Equation 16.11 becomes.

$\left[ {D - \left( {{K_1} + j{K_2}} \right)} \right]\,\,\left[ {D - {K_1} - j{K_2}} \right]i=0$

The solution for the above equation is

${i_c}=e{K_2}^t\left[ {{c_1}\,\cos \,{K_2}t + {c_2}\,\sin \,{K_2}t} \right]$

Therefore, the complete solution is

$i={i_c} + {i_p}$

$i=e{K_1}^t\left[ {{c_1}\,\cos \,{K_2}t + {c_2}\,\sin \,{K_2}t} \right]$

$+\frac{V}{{{{\sqrt{{R^2}+\left({\frac{1}{{\omega C}}-\omega L}\right)}}^2}}}\cos\,\left[{\omega t+\theta+{{\tan}^{-1}}\left({\frac{1}{{\omega CR}}-\frac{{\omega L}}{R}}\right)}\right]$

K2 becomes zero, when ${\left( {{R \over {2L}}} \right)^2}=1/LC$

Then the roots are equal which gives critically damped response.  Then, Eq.16.11 becomes       (D-K1) (D-K1) I = 0

The complementary function for the above equation is

${i_c}={e^{K1t}}\left( {{c_1} + {c_2}t} \right)$

Therefore, the complete solution is i=ic + ip

$i = eK_1^t\left[ {{c_1}+{c_2}t}\right]+\frac{V}{{\sqrt{{R^2}+}{{\left({\frac{1}{{\omega C}}-\omega L}\right)}^2}}}\cos\,\left[{\omega t+\theta+{{\tan }^{-1}}\left({\frac{1}{{\omega CR}}-\frac{{\omega L}}{R}}\right)}\right]$