## LESSON 15. Sinusoidal Response of R-L & R-C Circuit

15.1. Sinusoidal Response of R-L Circuit

Consider a circuit consisting of resistance and inductance as shown in Fig.15.1.  The switch, S, is closed at t=0.  At t=0, a sinusoidal voltage V cos (wt+q) is applied to the series R-L 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.

Fig.15.1

$V\,\cos \,\left( {\omega t + \theta } \right)=Ri + L{{di} \over {dt}}...................................................\left( {15.1} \right)$

${{di} \over {dt}} + {R \over L}i={V \over L}\,\,\cos \,\left( {\omega t + \theta } \right)$

The corresponding characteristic equation is

$\left( {D + {R \over L}} \right)i={{} \over L}\,\cos \,\left( {\omega t + \theta } \right)...................................................\left( {15.2} \right)$

For the above equation, the solution consists of two parts, viz. complementary function and particular integral.

The complementary function of the solution i is

${i_c}=c{e^{ - t\left( {R/L} \right)}}...................................................\left( {15.3} \right)$

The particular solution can be obtained by using undetermined co-efficient.

By assuming ${i_p}=A\,\cos \,\left( {\omega t + \theta } \right) + B\,\,\sin \,\left( {\omega t + \theta } \right)...................................................\left( {15.4} \right)$

$i_p^'=A\omega \,\sin \,\left( {\omega t + \theta } \right) + B\omega \,\cos \,\left( {\omega t + \theta } \right)...................................................\left( {15.5} \right)$

Substituting Eqs. 15.4 and 15.5 in Eq.15.2, we have

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

or $\left({ - A\omega + {{BR} \over L}} \right)\,\sin \,\left({\omega t+\theta}\right) + \left({B\omega + {{AR}\over L}}\right)\,\cos \,\,\left( {\omega t + \theta }\right)={V \over L}\,\cos \,\left({\omega t + \theta }\right)$

Comparing cosine terms and sine terms, we get

$-A\omega+{{BR}\over L}=0$

$B\omega+{{AR} \over L}={V \over L}$

From the above equations, we have

$A=V{R \over {{R^2} + {{\left( {\omega L} \right)}^2}}}$

$B=V{{\omega L} \over {{R^2} + {{\left( {\omega L} \right)}^2}}}$

Substituting the values of A and B in Eq.11.20, we get

${i_p}=V{R \over {{R^2} + {{\left( {\omega L} \right)}^2}}}\cos \,\left( {\omega t + \theta } \right) + V{{\omega L} \over {{R^2} + {{\left( {\omega L} \right)}^2}}}\,\sin \,\left( {\omega t + \theta } \right)...................................................\left( {15.6} \right)$

Putting $M\,\cos\,\phi={{VR}\over{{R^2}+{{\left({\omega L}\right)}^2}}}$

$M\,\sin\,\phi=\,V{{\omega L}\over{{R^2}+{{\left({\omega L} \right)}^2}}}$

to find M and ø, we divide one equation by the other

${{M\,\sin \,\phi} \over {M\,\cos \,\phi}}=\,\tan \,\phi \,=\,{{\omega L} \over R}\,$

Squaring both equation and adding, we get

${M^2}\,{\cos ^2}\,\phi+{M^2}\,{\sin ^2}\,\phi={{{V^2}} \over {{R^2} + {{\left( {\omega L} \right)}^2}}}$

or $M={V \over {\sqrt {{R^2} + {{\left( {\omega L} \right)}^2}} }}$

The particular current becomes

${i_p}={V\over{\sqrt{{R_2}+{{\left({\omega L}\right)}^2}}}}\,\cos\,\left({\omega t+\theta-{{\tan}^{-1}}{{\omega L}\over R}} \right)...................................................\left( {15.7} \right)$

The complete solution for the current $i={i_c} + {i_p}$

$i=c{e^{-t\left({R/L} \right)}} + {V \over{\sqrt{{R^2} + {{\left({\omega L} \right)}^2}}}}\,\cos \,\left({\omega t + \theta -{{\tan }^{-1}}{{\omega L} \over R}} \right)$

Since the inductor does not allow sudden changes in currens, at t=0, i=o

$c=-{V \over{\sqrt{{R^2} + {{\left( {\omega L}\right)}^2}} }}\,\,\cos \,\,\left( {\theta-{{\tan }^{-1}}{{\omega L}\over R}} \right)$

The complete solution for the current is

$i=\,{e^{-\left( {R/L} \right)t}}\left[ {{{-V} \over {\sqrt{{R^2} + {{\left( {\omega L} \right)}^2}} }}\,\,\cos \,\,\left({\theta-{{\tan }^{-1}}{{\omega L} \over R}} \right)} \right]$

$+{V \over{\sqrt{{R^2}+{{\left( {\omega L} \right)}^2}} }}\,\,\cos \,\,\left( {\omega t +\theta-{{\tan }^{-1}}{{\omega L}\over R}}\right)$

15.2. Sinusoidal Response of R-C Circuit

Consider a circuit consisting of resistance and capacitance in series as shown in Fig.15.2.  The switch, S, is closed at t=0.  At t=0, a sinusoidal voltage V cos (wt+q) is applied to the R-C circuit, where V is the amplitude of the wave and q is the phase angle.  Applying Kirchhoff’s voltage law to the circuit results in the following differential equation.

$V\,\cos \,\left( {\omega t + \theta } \right) = Ri + {1 \over C}\int {idt} ...................................................\left( {15.8} \right)$

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

$\left( {D + {1 \over {RC}}i=-{{V\omega } \over R}\,\sin \,\left( {\omega t + \theta } \right)} \right)...................................................\left( {15.9} \right)$

Fig. 15.2

The complementary function ic = ce-t/RC $...................................................\left( {15.10} \right)$

The particular solution can be obtained by using undermined coefficients.

${i_p}=A\,\cos \,\left( {\omega t + \theta } \right) + B\,\sin \,\left( {\omega t + \theta } \right)...................................................\left( {15.11} \right)$

$i_p^'=A\omega \,\sin \,\left( {\omega t + \theta } \right) + B\omega \,\cos \,\left( {\omega t + \theta } \right)...................................................\left( {15.12} \right)$

Substituting Eqs. 15.11 and 15.12 in Eq. 15.9, we get

$\left\{ { - A\omega \,\sin \,\left( {\omega t + \theta } \right) + B\omega \,\cos \,\left( {\omega t + \theta } \right)} \right\} + {1 \over {RC}}\left\{ {A\,\cos \,\left( {\omega t + \theta } \right) + B\,\sin \,\left( {\omega t + \theta } \right)} \right\}$

$=-{{V\omega } \over {R\,}}\,\sin \,\left( {\omega t + \theta } \right)$

Comparing both sides, $-A\omega+{B\over{RC}}=-{{V\omega}\over R}$

$B\omega+{A\over{RC}}=0$

From which,

$A = \frac{{VR}}{{{R^2} + {{\left( {\frac{1}{{\omega c}}} \right)}^2}}}$

and $B = \frac{{ - V}}{{\omega C\left[ {{R^2} + {{\left( {\frac{1}{{\omega c}}} \right)}^2}} \right]}}$

Substituting the values of A and B in Eq. 15.11, we have

${i_p} = \frac{{VR}}{{{R^2}+{{\left( {\frac{1}{{\omega c}}}\right)}^2}}}\,\cos\;\,\left( {\omega t+\theta }\right)+\frac{{-V}}{{\omega C\left[ {{R^2}+{{\left( {\frac{1}{{\omega C}}}\right)}^2}}\right]}}\,\sin\,\left( {\omega t+\theta } \right)$

Putting $M\,\cos\,\varphi=\frac{{VR}}{{{R^2}+{{\left({\frac{1}{{\omega C}}}\right)}^2}}}$

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

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

${{M\,\sin \,\phi } \over {M\,\cos \,\phi }}=\tan \,\phi={1 \over {\omega CR}}$

Squaring both equations and adding, we get

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

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

The particular current becomes

${i_p}=\frac{V}{{{R^2}+{{\left({\frac{1}{{\omega c}}}\right)}^2}}}\,\cos\;\left({\omega t+\theta+{{\tan}^{-1}}\frac{1}{{\omega CR}}}\right)...................................................\left( {15.13} \right)$

The complete solution for the current i=ic+ip

${i_p}=c{e^{-\left({t/RC}\right)}}+\frac{V}{{{R^2}+{{\left({\frac{1}{{\omega c}}}\right)}^2}}}\,\cos\;\left({\omega t+\theta+{{\tan}^{-1}}\frac{1}{{\omega CR}}}\right)...................................................\left( {15.14} \right)$

Since the capacitor does not allow sudden changes in voltages at $t=0,\,i={V \over R}\cos \,\theta$

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

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

The complete solution for the current is

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

$+\frac{V}{{\sqrt{{R^2}+{{\left({\frac{1}{{\omega C}}}\right)}^2}} }}\cos\,\left({\omega t+\theta+{{\tan}^{-1}}\frac{1}{{\omega CR}}}\right)...................................................\left( {15.15} \right)$